# Holomorphic Functions and the Cauchy-Riemann Equations: Why Complex Differentiability Is So Strong
> Complex differentiability asks for one limit from every direction. We derive the Cauchy-Riemann equations, show partial derivatives alone fail, and prove exp, sin and Log holomorphic.
> https://rikai.mugen-giken.com/en/mathematics/complex-analysis/holomorphic-functions
## 0. Key points
- The defining formula for the complex derivative is literally the same as in one real variable, but $h \to 0$ now happens in a plane (infinitely many directions), and the demand becomes incomparably stronger.
- The source of that strength is that the derivative is a single complex number rather than a $2 \times 2$ real matrix. A derivative in two real variables carries four real parameters; complex differentiability cuts this down to two. The two missing equations are the Cauchy-Riemann equations $u_x = v_y,\ u_y = -v_x$.
- Existence of the partial derivatives together with the equations does not imply complex differentiability ($f(x+iy) = \sqrt{|xy|}$ is a counterexample). Only when **total differentiability** of $u, v$ is added do we obtain a necessary and sufficient condition.
- In practice it suffices to check that $u, v$ are of class $C^1$ and satisfy the equations. Conversely, holomorphy forces $u, v$ to be of class $C^\infty$ automatically, but that is a deep fact resting on Cauchy's integral formula, proved in a later chapter.
- The functions $e^z$, $\cos z$, $\sin z$ are holomorphic on all of $\mathbb{C}$ (entire), while the logarithm has a principal branch $\operatorname{Log} z$, holomorphic on $\mathbb{C} \setminus (-\infty, 0]$ with $(\operatorname{Log} z)' = 1/z$.
---
## 1. Motivation: what happens if we transplant the definition of the derivative verbatim
For a real function, the derivative was defined as a limit of difference quotients (see [Limits and continuity ($\varepsilon$-$\delta$ arguments)](/en/mathematics/calculus/limits-and-continuity) and [The derivative: definition and basic rules](/en/mathematics/calculus/derivatives)).
$$
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}.
$$
Writing this formula requires only three things: (i) values can be subtracted, (ii) one can divide by the increment $h$, and (iii) there is a distance giving meaning to "$h \to 0$". The complex numbers have all three. Subtraction and division (for $h \ne 0$) are available, and the absolute value $|h|$ supplies a distance (see [Complex numbers and the complex plane](/en/mathematics/complex-analysis/complex-numbers)). So the formula above transfers to the complex world unchanged.
Yet although the shape of the definition is the same, its content is entirely different. On the real line there are essentially only two ways for $h$ to approach $0$: from the right and from the left. In the complex plane, $h$ may approach $0$ from every direction around the origin, and it may even spiral in. The demand that the difference quotient converge to one and the same complex number no matter how $h$ approaches is nothing like the demand in a world with two directions.
Another angle makes the difference even clearer. Identifying $\mathbb{C}$ with $\mathbb{R}^2$, a complex function $f$ is a map from the plane to the plane. In several-variable calculus (see [Differentiation in several variables and partial derivatives](/mathematics/calculus/multivariable-differentiation) and ), differentiability of $f$ meant that near $z_0$ the map $f$ can be approximated by a **real linear map** $\mathbb{R}^2 \to \mathbb{R}^2$. A real linear map is a $2 \times 2$ real matrix, hence four real parameters.
Complex differentiability, on the other hand, means that near $z_0$ the map $f$ can be approximated by **multiplication by a complex number $c$**, that is by $h \mapsto ch$. Multiplication by a complex number is also a real linear map, but it has only two real parameters, the real and imaginary parts of $c$. In other words, complex differentiability
> demands that the derivative lie in a $2$-dimensional subspace (the multiplications by complex numbers) inside a $4$-dimensional space (all real linear maps)
The $4 - 2 = 2$ dimensions that were lost reappear as extra equations. Those equations are the protagonists of this article: the Cauchy-Riemann equations.
Historically these equations were written down before either Cauchy or Riemann: d'Alembert and Euler already had them in the eighteenth century, in work on fluid dynamics and on mappings. Cauchy placed them at the foundation of the theory of complex integration in a series of papers from 1814 onward, and Riemann, in his 1851 dissertation, advanced the view that "it is this partial differential equation that determines a complex function". The present name is a later convention.
| | one real variable | two real variables | one complex variable |
|---|---|---|---|
| ways for $h \to 0$ | two, left and right | all directions in the plane | all directions in the plane |
| degrees of freedom of the derivative | one real number | $2\times 2$ matrix (four reals) | one complex number (two reals) |
| consequences of one derivative | continuity | continuity | infinitely differentiable, expandable in a power series |
The bottom right entry is the very reason the field of complex analysis exists. We prove that astonishing fact in [Cauchy's integral theorem and integral formula](/mathematics/complex-analysis/cauchy-integral-theorem) and [The powerful properties of holomorphic functions](/mathematics/complex-analysis/properties-of-holomorphic-functions). In this article our goal is the starting point of that road: a complete characterization of holomorphy in terms of real and imaginary parts.
---
## 2. Preliminaries: limits and continuity in the complex plane
Throughout we write $z = x + iy$ with $x, y \in \mathbb{R}$ and identify $\mathbb{C}$ with $\mathbb{R}^2$ via $z \leftrightarrow (x, y)$. The absolute value $|z| = \sqrt{x^2+y^2}$ is exactly the Euclidean norm, so the metric on $\mathbb{C}$ and the metric on $\mathbb{R}^2$ are the same, and the notions of open set, convergence and continuity coincide as well.
For $a \in \mathbb{C}$ and $r > 0$ we call $D(a, r) = \{\, z \in \mathbb{C} : |z - a| < r \,\}$ an **open disk**. A set $\Omega \subset \mathbb{C}$ is **open** if for each $a \in \Omega$ there is an $r > 0$ with $D(a, r) \subset \Omega$. A nonempty connected open set is called a **domain**. These are precisely the notions introduced in and .
Let $\Omega \subset \mathbb{C}$, $f : \Omega \to \mathbb{C}$, and let $a \in \mathbb{C}$ be an accumulation point of $\Omega$, that is, for every $r>0$ the set $\Omega \cap D(a,r)$ contains a point other than $a$. For a complex number $L$, if
$$
\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall z \in \Omega,\quad 0 < |z - a| < \delta \implies |f(z) - L| < \varepsilon
$$
holds, we write $\displaystyle\lim_{z \to a} f(z) = L$. If moreover $a \in \Omega$ and $\lim_{z\to a} f(z) = f(a)$, we say that $f$ is **continuous** at $a$.
As a string of symbols the definition is identical to the real one; only the meaning of $|\cdot|$ has changed to the complex modulus. Consequently the basic facts — uniqueness of limits, the limit laws for sums, differences, products and quotients, limits of composites, and "a composite of continuous functions is continuous" — carry over with proofs unchanged, word for word. Those proofs used only the triangle inequality $|\alpha + \beta| \le |\alpha| + |\beta|$ () and multiplicativity $|\alpha\beta| = |\alpha||\beta|$, both valid for complex numbers. We do not repeat them, and record only the splitting into real and imaginary parts.
Let $f = u + iv$, where $u = \operatorname{Re} f$ and $v = \operatorname{Im} f$ are real valued, and let $L = A + iB$ with $A, B \in \mathbb{R}$. If $a$ is an accumulation point of $\Omega$, then
$$
\lim_{z \to a} f(z) = L
\quad\Longleftrightarrow\quad
\lim_{(x,y)\to(\alpha,\beta)} u(x,y) = A \ \text{ and } \ \lim_{(x,y)\to(\alpha,\beta)} v(x,y) = B,
$$
where $a = \alpha + i\beta$.
We use the relation between modulus and components,
$$
\max\bigl(|u - A|,\ |v - B|\bigr) \ \le\ |f - L| \ \le\ |u - A| + |v - B| .
$$
The left inequality follows from $|\operatorname{Re} w| \le |w|$ and $|\operatorname{Im} w| \le |w|$ applied to $w = f - L$; the right one follows at once from the triangle inequality together with $|i(v-B)| = |v-B|$.
($\Rightarrow$) Let $\varepsilon > 0$ be arbitrary. By hypothesis there is $\delta > 0$ such that $0 < |z-a| < \delta$ implies $|f(z) - L| < \varepsilon$. The left inequality then gives $|u - A| \le |f - L| < \varepsilon$ and $|v - B| < \varepsilon$ simultaneously. Since $|z - a|$ equals the Euclidean distance between $(x,y)$ and $(\alpha,\beta)$, this says precisely that $u \to A$ and $v \to B$.
($\Leftarrow$) Let $\varepsilon > 0$. By hypothesis there are $\delta_1, \delta_2 > 0$ such that $0 < |z - a| < \delta_1$ gives $|u - A| < \varepsilon/2$ and $0 < |z-a| < \delta_2$ gives $|v - B| < \varepsilon/2$. Taking $\delta = \min(\delta_1, \delta_2)$, for $0 < |z-a| < \delta$ the right inequality yields $|f - L| < \varepsilon/2 + \varepsilon/2 = \varepsilon$.
The proposition looks modest, but we invoke it every time we pass between one complex limit and two real limits.
---
## 3. Complex differentiability and holomorphy
Let $\Omega \subset \mathbb{C}$ be open, $f : \Omega \to \mathbb{C}$ and $z_0 \in \Omega$. If the limit
$$
\lim_{h \to 0} \frac{f(z_0 + h) - f(z_0)}{h}
$$
exists, where $h$ ranges over nonzero complex numbers, we say $f$ is **complex differentiable** at $z_0$ and write $f'(z_0)$ for the limit.
Since $\Omega$ is open, $z_0 + h \in \Omega$ for all sufficiently small $|h|$, so the difference quotient is defined.
If $f$ is complex differentiable at **every point** of an open set $\Omega$, we say that $f$ is **holomorphic** on $\Omega$. For a point $z_0$, the statement "$f$ is holomorphic at $z_0$" means that $f$ is holomorphic on some open neighborhood of $z_0$. A function holomorphic on all of $\mathbb{C}$ is called **entire**.
Rewriting the definition in terms of "first-order approximation" rather than a limit makes the later arguments easier.
That $f$ is complex differentiable at $z_0$ with $f'(z_0) = c$ is equivalent to the following: there exist a complex number $c$ and a function $\varepsilon$ with $\varepsilon(h) \to 0$ as $h \to 0$ such that
$$
f(z_0 + h) = f(z_0) + c\,h + \varepsilon(h)\,|h| \qquad (\varepsilon(0) = 0)
$$
holds for all sufficiently small $|h|$.
($\Rightarrow$) For $h \ne 0$ set
$$
\varepsilon(h) := \frac{1}{|h|}\Bigl( f(z_0+h) - f(z_0) - c\,h \Bigr)
$$
and $\varepsilon(0) = 0$. Then the identity holds by definition, and it remains to show $\varepsilon(h) \to 0$. Now
$$
|\varepsilon(h)| = \frac{|f(z_0+h) - f(z_0) - ch|}{|h|} = \left| \frac{f(z_0+h)-f(z_0)}{h} - c \right|
$$
(dividing by $|h|$ outside is the same as dividing by $h$ inside the modulus), and the right-hand side tends to $0$ as $h \to 0$ by complex differentiability.
($\Leftarrow$) Conversely, if the identity holds then for $h \ne 0$
$$
\frac{f(z_0+h)-f(z_0)}{h} - c = \frac{\varepsilon(h)|h|}{h}, \qquad \left|\frac{\varepsilon(h)|h|}{h}\right| = |\varepsilon(h)| \to 0,
$$
so the difference quotient converges to $c$.
If $f$ is complex differentiable at $z_0$, then $f$ is continuous at $z_0$.
Let $h \to 0$ in the representation $f(z_0+h) = f(z_0) + f'(z_0)h + \varepsilon(h)|h|$ from . We have $|f'(z_0) h| = |f'(z_0)||h| \to 0$, and $|\varepsilon(h)|h|| = |\varepsilon(h)|\,|h| \to 0$ since $\varepsilon(h) \to 0$ and $|h| \to 0$. Hence $f(z_0+h) \to f(z_0)$, that is, $f$ is continuous at $z_0$.
Let $f, g$ be complex differentiable at $z_0$. Then the following hold.
1. For all $\alpha, \beta \in \mathbb{C}$, the function $\alpha f + \beta g$ is complex differentiable at $z_0$ with $(\alpha f + \beta g)'(z_0) = \alpha f'(z_0) + \beta g'(z_0)$.
2. $fg$ is complex differentiable at $z_0$ with $(fg)'(z_0) = f'(z_0)g(z_0) + f(z_0)g'(z_0)$.
3. If $g(z_0) \ne 0$, then $f/g$ is defined on a neighborhood of $z_0$, is complex differentiable at $z_0$, and
$$
\left(\frac{f}{g}\right)'(z_0) = \frac{f'(z_0)g(z_0) - f(z_0)g'(z_0)}{g(z_0)^2}.
$$
4. If moreover $g$ is defined on a neighborhood of $f(z_0)$ and complex differentiable at $f(z_0)$, then $g \circ f$ is complex differentiable at $z_0$ with $(g\circ f)'(z_0) = g'(f(z_0))\, f'(z_0)$.
The proofs are exactly the computations of the one-variable real case. They use only the field axioms and the limit laws, and nothing changes when $\mathbb{R}$ is replaced by $\mathbb{C}$. We write out 2 and 4 as representatives.
**2 (product).** For $h \ne 0$,
$$
\frac{f(z_0+h)g(z_0+h) - f(z_0)g(z_0)}{h}
= \frac{f(z_0+h)-f(z_0)}{h}\,g(z_0+h) + f(z_0)\,\frac{g(z_0+h)-g(z_0)}{h}.
$$
This is an identity obtained by adding and subtracting $f(z_0)g(z_0+h)$ in the numerator. Letting $h \to 0$, the first term converges to $f'(z_0)g(z_0)$ (because $g(z_0+h) \to g(z_0)$ by ) and the second to $f(z_0)g'(z_0)$.
**4 (composition).** Here is the cleanest tool. Put $w_0 = f(z_0)$, $c = f'(z_0)$, $d = g'(w_0)$, so that
$$
f(z_0 + h) = w_0 + ch + \varepsilon_1(h)|h|, \qquad
g(w_0 + k) = g(w_0) + dk + \varepsilon_2(k)|k|
$$
with $\varepsilon_1(h) \to 0$ as $h\to0$ and $\varepsilon_2(k)\to 0$ as $k \to 0$. Setting $k = k(h) := ch + \varepsilon_1(h)|h|$ we get $|k| \le (|c| + |\varepsilon_1(h)|)|h|$, so in particular $k \to 0$ as $h \to 0$. Substituting,
$$
g(f(z_0+h)) = g(w_0) + d\bigl(ch + \varepsilon_1(h)|h|\bigr) + \varepsilon_2(k)|k|
= g(w_0) + (dc)h + \eta(h)|h|,
$$
where $\eta(h) := d\,\varepsilon_1(h) + \varepsilon_2(k(h))\cdot \dfrac{|k(h)|}{|h|}$ for $h \ne 0$. Since $|k(h)|/|h| \le |c| + |\varepsilon_1(h)|$ is bounded for small $h$, and $\varepsilon_2(k(h)) \to 0$, $\varepsilon_1(h)\to 0$, we get $\eta(h) \to 0$. Applying once more gives $(g\circ f)'(z_0) = dc$.
Parts 1 and 3 are obtained in the same way, by reading the real one-variable proofs with $\mathbb{R}$ replaced by $\mathbb{C}$.
For $n \in \mathbb{N}$ let $f(z) = z^n$. Then $f$ is entire with $f'(z) = nz^{n-1}$. Indeed, by the binomial theorem
$$
(z_0+h)^n = z_0^n + n z_0^{n-1} h + \sum_{k=2}^{n} \binom{n}{k} z_0^{\,n-k} h^k,
$$
so for $h \ne 0$
$$
\frac{(z_0+h)^n - z_0^n}{h} = n z_0^{n-1} + h \sum_{k=2}^{n} \binom{n}{k} z_0^{\,n-k} h^{k-2}.
$$
For $|h| \le 1$ the sum in the second term is bounded by $\sum_{k=2}^n \binom{n}{k}|z_0|^{n-k}$, so the whole second term tends to $0$ as $h \to 0$ thanks to the factor $h$. Hence the limit is $nz_0^{n-1}$.
Combining with part 1 of , every polynomial $P(z) = a_0 + a_1 z + \cdots + a_n z^n$ with $a_k \in \mathbb{C}$ is entire, with $P'(z) = a_1 + 2a_2 z + \cdots + n a_n z^{n-1}$.
Consider $f(z) = \bar{z}$. Fix $z_0$ arbitrarily; for $h \ne 0$,
$$
\frac{f(z_0+h) - f(z_0)}{h} = \frac{\overline{z_0 + h} - \bar{z_0}}{h} = \frac{\bar{h}}{h}.
$$
Try two ways of letting $h \to 0$. With $h = t$ ($t$ a nonzero real), $\bar h / h = t/t = 1$, so the limit as $t \to 0$ is $1$. With $h = it$ ($t$ a nonzero real), $\bar h / h = \overline{it}/(it) = (-it)/(it) = -1$, so the limit is $-1$. As $1 \ne -1$, the limit in does not exist. Since $z_0$ was arbitrary, $\bar z$ is complex differentiable at no point of the plane.
As a map of $\mathbb{R}^2$, however, $\bar z$ is the linear map $(x,y) \mapsto (x, -y)$, reflection in the real axis, which is as smooth as one could wish. Smoothness is not what matters; **whether orientation is preserved** is. That is the first warning.
Let $f(z) = |z|^2$. Using $|z|^2 = z\bar z$ we compute the difference quotient. For $h \ne 0$,
$$
\frac{|z_0+h|^2 - |z_0|^2}{h}
= \frac{(z_0+h)(\bar{z_0}+\bar h) - z_0 \bar{z_0}}{h}
= \frac{z_0 \bar h + h \bar{z_0} + h\bar h}{h}
= z_0 \frac{\bar h}{h} + \bar{z_0} + \bar h .
$$
As $h \to 0$ we have $\bar h \to 0$, but as seen in the quotient $\bar h / h$ has no limit. Hence for $z_0 \ne 0$ the first term oscillates and the limit fails to exist. For $z_0 = 0$, on the other hand, the whole expression equals $\bar h$, which tends to $0$ as $h \to 0$.
Thus $f(z) = |z|^2$ is complex differentiable only at $z_0 = 0$, with $f'(0) = 0$, and it is **holomorphic at no point**: every neighborhood of $0$ contains points other than $0$, where $f$ is not differentiable.
---
## 4. The Cauchy-Riemann equations
In and we derived a contradiction by comparing two ways of approaching, along the real axis and along the imaginary axis. Carrying out this comparison for a general $f$ produces a system of partial differential equations that the real and imaginary parts must satisfy.
From now on we split $f = u + iv$ and regard $u, v$ as real-valued functions of $(x,y)$. Partial derivatives are abbreviated as $u_x = \partial u/\partial x$ and so on.
Let $f = u + iv$ be defined on an open set $\Omega \subset \mathbb{C}$ and complex differentiable at a point $z_0 = x_0 + i y_0 \in \Omega$. Then:
1. $u$ and $v$ are totally differentiable at $(x_0, y_0)$ in the sense of two real variables;
2. their partial derivatives satisfy the **Cauchy-Riemann equations**
$$
u_x(x_0,y_0) = v_y(x_0,y_0), \qquad u_y(x_0,y_0) = -\,v_x(x_0,y_0);
$$
3. the derivative is given by
$$
f'(z_0) = u_x(x_0,y_0) + i\,v_x(x_0,y_0) = v_y(x_0,y_0) - i\,u_y(x_0,y_0).
$$
**Step 1: comparing limits along two directions gives 2 and 3.**
The limit in holds for every way of letting $h \to 0$, so in particular the limits with $h$ restricted to the real axis or to the imaginary axis both equal $f'(z_0)$.
First take $h = t$ with $t \in \mathbb{R}$, $t \ne 0$:
$$
\frac{f(z_0+t)-f(z_0)}{t}
= \frac{u(x_0+t, y_0) - u(x_0,y_0)}{t} + i\,\frac{v(x_0+t,y_0)-v(x_0,y_0)}{t}.
$$
The left-hand side converges to $f'(z_0)$ as $t \to 0$. The real and imaginary parts on the right are real valued, so by each of their limits exists separately and equals $\operatorname{Re} f'(z_0)$ and $\operatorname{Im} f'(z_0)$ respectively. Those limits are precisely the definitions of the partial derivatives, so $u_x(x_0,y_0)$ and $v_x(x_0,y_0)$ exist and
$$
f'(z_0) = u_x(x_0,y_0) + i\,v_x(x_0,y_0).
$$
Call this the representation along the real direction.
Next take $h = it$ with $t \in \mathbb{R}$, $t \ne 0$. Using $1/i = -i$,
$$
\frac{f(z_0+it)-f(z_0)}{it}
= -i\left[\frac{u(x_0, y_0+t)-u(x_0,y_0)}{t} + i\,\frac{v(x_0,y_0+t)-v(x_0,y_0)}{t}\right].
$$
The bracket converges as $t \to 0$ to $u_y(x_0,y_0) + i v_y(x_0,y_0)$ (again by , which also yields the existence of these partial derivatives). Hence
$$
f'(z_0) = -i\bigl(u_y + i v_y\bigr) = v_y(x_0,y_0) - i\,u_y(x_0,y_0),
$$
the representation along the imaginary direction. These are two representations of the same complex number $f'(z_0)$, so comparing real parts and imaginary parts gives
$$
u_x = v_y, \qquad v_x = -u_y ,
$$
which are the Cauchy-Riemann equations. The two formulas in assertion 3 are exactly the two representations just obtained.
**Step 2: proof of 1.**
Write $c = f'(z_0) = a + ib$ with $a, b \in \mathbb{R}$, and $h = s + it$ with $s,t\in\mathbb{R}$. By ,
$$
f(z_0+h) - f(z_0) = c\,h + \varepsilon(h)|h|, \qquad \varepsilon(h) \to 0 \ (h\to 0).
$$
Here $c h = (a+ib)(s+it) = (as - bt) + i(bs + at)$. Taking real and imaginary parts of both sides and writing $\varepsilon = \varepsilon_1 + i\varepsilon_2$ with $\varepsilon_1,\varepsilon_2$ real valued, we get
$$
\begin{aligned}
u(x_0+s, y_0+t) - u(x_0,y_0) &= a s - b t + \varepsilon_1(h)\sqrt{s^2+t^2},\\
v(x_0+s, y_0+t) - v(x_0,y_0) &= b s + a t + \varepsilon_2(h)\sqrt{s^2+t^2}.
\end{aligned}
$$
Here $|h| = \sqrt{s^2+t^2}$ and $|\varepsilon_j(h)| \le |\varepsilon(h)| \to 0$. These two identities say exactly that $u$ is approximated to first order by the linear map $(s,t)\mapsto as - bt$ and $v$ by $(s,t)\mapsto bs+at$, that is, that $u, v$ are totally differentiable at $(x_0,y_0)$. As a bonus we can read off the gradients: $(u_x, u_y) = (a, -b)$ and $(v_x, v_y) = (b, a)$, which confirms the conclusion of Step 1 once more.
For $f(z) = z^2$ we have $f(x+iy) = (x+iy)^2 = (x^2 - y^2) + i\,(2xy)$, so $u = x^2 - y^2$ and $v = 2xy$. The partial derivatives are
$$
u_x = 2x,\quad u_y = -2y,\quad v_x = 2y,\quad v_y = 2x .
$$
Indeed $u_x = 2x = v_y$ and $u_y = -2y = -(2y) = -v_x$, so the Cauchy-Riemann equations hold on the whole plane. Part 3 of gives
$$
f'(z) = u_x + i v_x = 2x + i\,2y = 2(x+iy) = 2z,
$$
in agreement with .
For $f(z) = \bar z$ from we have $u = x$ and $v = -y$. Since $u_x = 1$ and $v_y = -1$, the identity $u_x = v_y$ holds at no point, because $1 \ne -1$. By the contrapositive of , $f$ is complex differentiable nowhere. The computation with difference quotients in has been replaced by a mechanical check of partial derivatives.
Similarly $f(z) = |z|^2 = x^2+y^2$ has $u = x^2+y^2$, $v = 0$, hence $u_x = 2x$, $v_y = 0$, $u_y = 2y$, $-v_x = 0$. The equations hold only when $x = y = 0$, that is, only at the origin, consistent with the conclusion of .
### 4.1. Wirtinger derivatives: "not depending on $\bar z$"
The Cauchy-Riemann equations are two real identities, but packaging them into a single complex identity makes their meaning plain.
When $u, v$ are partially differentiable, define for $f = u+iv$
$$
\frac{\partial f}{\partial z} := \frac{1}{2}\left( \frac{\partial f}{\partial x} - i \frac{\partial f}{\partial y}\right),
\qquad
\frac{\partial f}{\partial \bar z} := \frac{1}{2}\left( \frac{\partial f}{\partial x} + i \frac{\partial f}{\partial y}\right),
$$
where $\partial f/\partial x = u_x + i v_x$ and $\partial f/\partial y = u_y + i v_y$.
With the notation above, at the point $(x_0,y_0)$,
$$
\text{the Cauchy-Riemann equations} \iff \frac{\partial f}{\partial \bar z}(z_0) = 0,
$$
and in that case, if $f'(z_0)$ exists, then $f'(z_0) = \dfrac{\partial f}{\partial z}(z_0)$.
We compute directly from the definition:
$$
\frac{\partial f}{\partial \bar z}
= \frac{1}{2}\Bigl[(u_x + iv_x) + i(u_y + iv_y)\Bigr]
= \frac{1}{2}\Bigl[(u_x - v_y) + i\,(v_x + u_y)\Bigr].
$$
A complex number vanishes exactly when its real and imaginary parts both vanish, so $\partial f/\partial \bar z = 0$ is equivalent to $u_x - v_y = 0$ and $v_x + u_y = 0$, that is, to $u_x = v_y$ and $u_y = -v_x$.
Similarly
$$
\frac{\partial f}{\partial z}
= \frac{1}{2}\Bigl[(u_x + iv_x) - i(u_y + iv_y)\Bigr]
= \frac{1}{2}\Bigl[(u_x + v_y) + i\,(v_x - u_y)\Bigr],
$$
and under the equations $u_x + v_y = 2u_x$ and $v_x - u_y = 2v_x$, so $\partial f/\partial z = u_x + i v_x$. By part 3 of this equals $f'(z_0)$.
### 4.2. The equations alone are not enough
is a one-way statement: complex differentiability $\Rightarrow$ existence of the partial derivatives plus the equations. The converse fails.
Let
$$
f(x+iy) = \sqrt{|xy|},
$$
which is real valued, so $u = \sqrt{|xy|}$ and $v \equiv 0$.
**(Continuity.)** By the inequality of arithmetic and geometric means, $|xy| \le (x^2+y^2)/2$, hence $0 \le u \le |z|/\sqrt{2}$. Therefore $u \to 0 = u(0,0)$ as $z \to 0$, and $f$ is continuous at the origin. Away from the origin, $u$ is the composite of the continuous map $(x,y)\mapsto |xy|$ with the continuous map $s \mapsto \sqrt{s}$ on $s \ge 0$, hence continuous.
**(Partial derivatives at the origin and the equations.)** Since $u(t, 0) = \sqrt{|t\cdot 0|} = 0$ for **every** $t$,
$$
u_x(0,0) = \lim_{t\to 0}\frac{u(t,0)-u(0,0)}{t} = \lim_{t\to0}\frac{0-0}{t} = 0 .
$$
Likewise $u(0,t) = 0$ gives $u_y(0,0) = 0$, and $v \equiv 0$ gives $v_x(0,0) = v_y(0,0) = 0$. So at the origin $u_x = 0 = v_y$ and $u_y = 0 = -v_x$: the Cauchy-Riemann equations hold.
**(Failure of complex differentiability.)** For $h = s + it \ne 0$ the difference quotient is
$$
\frac{f(h) - f(0)}{h} = \frac{\sqrt{|st|}}{s+it}.
$$
Along $h = t$ (the real direction, $t\ne0$) it equals $\sqrt{|t\cdot 0|}/t = 0$. Along $h = t(1+i)$ with $t > 0$, however, $s = t$, so
$$
\frac{\sqrt{t^2}}{t(1+i)} = \frac{t}{t(1+i)} = \frac{1}{1+i} = \frac{1-i}{2} \ne 0 .
$$
Two ways of approaching give different limits, so the complex derivative at the origin does not exist.
**(What goes wrong.)** In the light of part 1 of , what fails is the total differentiability of $u$. If $u$ were totally differentiable at the origin, its gradient would be $(u_x, u_y) = (0,0)$, forcing $u(s,t) = o(\sqrt{s^2+t^2})$. But along $s = t > 0$,
$$
\frac{u(t,t)}{\sqrt{t^2+t^2}} = \frac{t}{\sqrt{2}\,t} = \frac{1}{\sqrt 2},
$$
which does not tend to $0$. Existence of the partial derivatives is information about two directions only; total differentiability is a condition treating all directions uniformly. That basic caution from multivariable calculus is exactly what bites here.
Set $f(0) = 0$ and $f(z) = e^{-z^{-4}}$ for $z \ne 0$. On both the real and the imaginary axis the values of $f$ are the real numbers $e^{-1/t^4}$, and $e^{-1/t^4}/t \to 0$ as $t\to0$, so all four partial derivatives at the origin vanish and the equations hold. Yet along $z = t e^{i\pi/4}$ we have $z^4 = -t^4$, hence $f(z) = e^{1/t^4} \to \infty$, so $f$ is not even continuous at the origin. As long as the equations are imposed only at a point, matters can be made as bad as one likes.
---
## 5. Criteria for holomorphy
What was missing was total differentiability. Supplying it, the converse direction holds.
Let $f = u+iv$ be defined on an open set $\Omega$ and let $z_0 = x_0+iy_0 \in \Omega$. Suppose $u$ and $v$ are totally differentiable at $(x_0,y_0)$ and satisfy the Cauchy-Riemann equations
$$
u_x(x_0,y_0) = v_y(x_0,y_0), \qquad u_y(x_0,y_0) = -\,v_x(x_0,y_0)
$$
there. Then $f$ is complex differentiable at $z_0$, with $f'(z_0) = u_x(x_0,y_0) + i\,v_x(x_0,y_0)$.
Put $a := u_x(x_0,y_0)$ and $b := v_x(x_0,y_0)$. The equations give $v_y = a$ and $u_y = -b$.
Total differentiability of $u$ and $v$ at $(x_0,y_0)$ means that as $h = s+it \to 0$,
$$
\begin{aligned}
u(x_0+s,y_0+t) - u(x_0,y_0) &= u_x s + u_y t + \varepsilon_1(h)|h| = a s - b t + \varepsilon_1(h)|h|,\\
v(x_0+s,y_0+t) - v(x_0,y_0) &= v_x s + v_y t + \varepsilon_2(h)|h| = b s + a t + \varepsilon_2(h)|h|
\end{aligned}
$$
with $\varepsilon_1(h), \varepsilon_2(h) \to 0$ and $|h| = \sqrt{s^2+t^2}$. In the second equality of each line we used the Cauchy-Riemann equations.
Since $f = u + iv$, combining the two lines with $1$ and $i$ gives
$$
f(z_0+h) - f(z_0) = (as - bt) + i\,(bs + at) + \bigl(\varepsilon_1(h) + i\varepsilon_2(h)\bigr)|h| .
$$
The key is the following factorization. Setting $c := a + ib$,
$$
c\,h = (a+ib)(s+it) = (as - bt) + i\,(bs + at),
$$
whose right-hand side matches exactly the first two terms above. That is,
$$
f(z_0+h) - f(z_0) = c\,h + \varepsilon(h)|h|, \qquad \varepsilon(h) := \varepsilon_1(h) + i\,\varepsilon_2(h) \to 0 .
$$
This is precisely the condition of , so $f$ is complex differentiable at $z_0$ with $f'(z_0) = c = u_x + i v_x$.
It is worth noting where the equations were used. Without them the Jacobian matrix would be a general real matrix $\begin{pmatrix} u_x & u_y \\ v_x & v_y\end{pmatrix}$, which cannot be written as multiplication by a complex number $c$. The equations were used solely to force this matrix into the shape $\begin{pmatrix} a & -b \\ b & a\end{pmatrix}$, **the shape of multiplication by a complex number**.
For $f = u+iv$ on an open set $\Omega$ and $z_0 = x_0+iy_0 \in \Omega$, the following are equivalent.
1. $f$ is complex differentiable at $z_0$.
2. $u$ and $v$ are totally differentiable at $(x_0,y_0)$ and satisfy the Cauchy-Riemann equations there.
1 $\Rightarrow$ 2 is parts 1 and 2 of ; 2 $\Rightarrow$ 1 is .
Given a concrete function, checking total differentiability directly is tedious. In practice one goes through the standard sufficient condition of multivariable calculus: continuous partial derivatives imply total differentiability.
Let $\Omega \subset \mathbb{C}$ be open and $f = u + iv$. If $u, v$ are of class $C^1$ on $\Omega$ (the four partial derivatives $u_x, u_y, v_x, v_y$ exist and are continuous on $\Omega$) and satisfy the Cauchy-Riemann equations at every point of $\Omega$, then $f$ is holomorphic on $\Omega$ with $f' = u_x + i v_x$.
By a theorem of multivariable calculus, if the partial derivatives exist on a neighborhood of a point $(x_0,y_0)$ and are continuous at that point, then the function is totally differentiable there (). By hypothesis $u, v$ satisfy this at every point of $\Omega$, hence are totally differentiable at every point. Applying at each point shows that $f$ is complex differentiable at every point of $\Omega$, that is, holomorphic on $\Omega$ by .
|"continuity of the partials"| B["u, v totally differentiable and satisfying CR"]
B -->|"sufficiency theorem"| C["f complex differentiable at z0"]
C -->|"necessity theorem"| B
C -->|"necessity theorem"| D["u, v partially differentiable and satisfying CR"]
D -.->|"counterexample exists"| C`} />
Reading the diagram from top to bottom gives the practical test. As for the reverse directions, one can go back from "complex differentiable" to "totally differentiable plus the equations" (), but not from "partially differentiable plus the equations" to "complex differentiable" (). This asymmetry is what separates the next two statements.
Let $\Omega \subset \mathbb{C}$ be open and $f = u+iv : \Omega \to \mathbb{C}$. The following are equivalent.
1. $f$ is holomorphic on $\Omega$.
2. $u, v$ are of class $C^1$ on $\Omega$ and satisfy the Cauchy-Riemann equations at every point of $\Omega$.
2 $\Rightarrow$ 1 is .
For 1 $\Rightarrow$ 2, the pointwise validity of the equations is immediate from . What remains is continuity of the partial derivatives, and this cannot be proved with the tools of this article alone. We use the following fact: **a function holomorphic on $\Omega$ is infinitely often complex differentiable on $\Omega$** (). It is derived from Goursat's theorem and Cauchy's integral formula, and is proved in [Cauchy's integral theorem and integral formula](/mathematics/complex-analysis/cauchy-integral-theorem). Granting it, $f'$ is itself holomorphic on $\Omega$ and in particular continuous by . By part 3 of we have $u_x = \operatorname{Re} f'$, $v_x = \operatorname{Im} f'$, $v_y = \operatorname{Re} f'$ and $u_y = -\operatorname{Im} f'$, so all four partial derivatives are continuous, being real and imaginary parts (possibly with a sign) of the continuous function $f'$. Repeating the argument in fact gives $u, v \in C^\infty$.
It is worth being aware that 1 $\Rightarrow$ 2 in depends on a later chapter. The proof of the integral theorem (Goursat's argument) uses only the definition of holomorphy, namely pointwise complex differentiability, and does not assume $C^1$ regularity, so there is no circularity. How far the hypotheses can be weakened is discussed in the Appendix.
### 5.1. The geometry behind the equations: rotation and scaling
Let $f = u+iv$ be complex differentiable at $z_0$, and let
$$
J_f(x_0,y_0) = \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}
$$
be the Jacobian matrix of $f$ viewed as a map $\mathbb{R}^2 \to \mathbb{R}^2$ (all entries evaluated at $(x_0,y_0)$). Then, with $a = u_x$ and $b = v_x$,
$$
J_f(x_0,y_0) = \begin{pmatrix} a & -b \\ b & a \end{pmatrix}
= |f'(z_0)| \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}
\quad (f'(z_0) \ne 0,\ \theta = \arg f'(z_0)),
$$
and in particular $\det J_f(x_0,y_0) = a^2 + b^2 = |f'(z_0)|^2 \ge 0$.
By we have $u_x = v_y = a$ and $u_y = -v_x = -b$, so the matrix takes the form $\begin{pmatrix} a & -b \\ b & a\end{pmatrix}$. Writing $f'(z_0) = a + ib$ in polar form $f'(z_0) = r e^{i\theta}$ with $r = |f'(z_0)| = \sqrt{a^2+b^2}$ gives $a = r\cos\theta$ and $b = r\sin\theta$, so the matrix is $r$ times the rotation matrix through the angle $\theta$. The determinant is $a\cdot a - (-b)\cdot b = a^2+b^2$.
When $f'(z_0) \ne 0$, the first-order approximation of $f$ at $z_0$ is the map "rotate by $\theta$ and scale by $r$". Rotations and dilations preserve angles, so the angle between two curves meeting at $z_0$ is preserved (orientation included) by $f$. This is **conformality**, the starting point of [Conformal mappings and the Riemann mapping theorem](/mathematics/complex-analysis/conformal-mapping). We can now also see why $\bar z$ in was not complex differentiable: its Jacobian matrix $\begin{pmatrix}1 & 0\\ 0& -1\end{pmatrix}$ has determinant $-1$, so the map reverses orientation.
If $f = u + iv$ is holomorphic on an open set $\Omega$, then $u, v$ are of class $C^\infty$ on $\Omega$ and satisfy Laplace's equation
$$
\Delta u = u_{xx} + u_{yy} = 0, \qquad \Delta v = v_{xx} + v_{yy} = 0 ;
$$
that is, $u$ and $v$ are harmonic.
That $u, v \in C^\infty(\Omega)$ follows from the fact quoted in the proof of , that holomorphic functions are infinitely often complex differentiable. Being of class $C^2$ allows us to interchange the order of partial differentiation (). Differentiating the Cauchy-Riemann equation $u_x = v_y$ with respect to $x$ gives $u_{xx} = v_{yx}$, and differentiating $u_y = -v_x$ with respect to $y$ gives $u_{yy} = -v_{xy}$. Adding,
$$
u_{xx} + u_{yy} = v_{yx} - v_{xy} = 0,
$$
the last equality using $v_{yx} = v_{xy}$ for $v \in C^2$. For $v$, differentiate $u_x = v_y$ with respect to $y$ and $u_y = -v_x$ with respect to $x$ to get $u_{xy} = v_{yy}$ and $u_{yx} = -v_{xx}$; subtracting yields $v_{yy} + v_{xx} = u_{xy} - u_{yx} = 0$.
This corollary is why complex analysis applies to two-dimensional problems governed by Laplace's equation: electrostatic fields, steady heat conduction, incompressible irrotational flow. Conversely, for a harmonic function $u$ on a simply connected domain one can construct a harmonic conjugate $v$ making $u + iv$ holomorphic; we carry out that procedure in .
Let $\Omega$ be a domain (a connected open set) and let $f$ be holomorphic on $\Omega$ with $f'(z) = 0$ for all $z \in \Omega$. Then $f$ is constant on $\Omega$.
By part 3 of we have $u_x = \operatorname{Re} f' = 0$ and $v_x = \operatorname{Im} f' = 0$ on $\Omega$, and the equations then give $v_y = u_x = 0$ and $u_y = -v_x = 0$ as well. So all four partial derivatives of $u, v$ vanish identically on $\Omega$. Moreover, by part 1 of , $u$ and $v$ are totally differentiable at every point.
Fix $a \in \Omega$ and set $S = \{\, z \in \Omega : f(z) = f(a) \,\}$.
$S$ is nonempty, since $a \in S$. As $f$ is continuous (), $S = f^{-1}(\{f(a)\})$ is relatively closed in $\Omega$.
We show $S$ is open. Take $z_0 \in S$ and choose $r>0$ with $D(z_0, r) \subset \Omega$. For $z \in D(z_0,r)$, the disk is convex, so the segment $\gamma(t) = z_0 + t(z - z_0)$, $t \in [0,1]$, lies in $D(z_0,r)$. Put $g(t) := u(\gamma(t))$. Since $u$ is totally differentiable, the chain rule applies and
$$
g'(t) = u_x(\gamma(t))\,\operatorname{Re}(z-z_0) + u_y(\gamma(t))\,\operatorname{Im}(z-z_0) = 0
$$
because $u_x = u_y = 0$. Thus $g$ is continuous on $[0,1]$ with vanishing derivative on $(0,1)$, so by the mean value theorem () $g(1) = g(0)$, that is, $u(z) = u(z_0)$. The same argument for $v$ gives $v(z) = v(z_0)$. Hence $f(z) = f(z_0) = f(a)$ and $D(z_0,r) \subset S$.
Since $\Omega$ is connected and $S$ is a nonempty subset of $\Omega$ that is both open and closed, $S = \Omega$. So $f$ is constantly equal to $f(a)$ on $\Omega$.
Suppose $f$ is holomorphic on a domain $\Omega$ and $u = \operatorname{Re} f$ is constant. Then $u_x = u_y = 0$ on $\Omega$, and the Cauchy-Riemann equations give $v_y = u_x = 0$ and $v_x = -u_y = 0$. By part 3 of , $f' = u_x + iv_x = 0$, so $f$ is constant by .
Nothing of the kind happens for two real variables. For instance $F(x,y) = (0,\ x^2 + y^3)$ is a $C^\infty$ map $\mathbb{R}^2 \to \mathbb{R}^2$ whose first component is constant while the second is not: the two real components can be chosen completely independently. The moment holomorphy is imposed, the imaginary part is entirely pinned down by the real part, up to an additive constant. This rigidity leads to the identity theorem and the maximum principle, treated in [The powerful properties of holomorphic functions](/mathematics/complex-analysis/properties-of-holomorphic-functions).
---
## 6. Holomorphy of the basic complex functions
By , polynomials are entire. Let $P, Q$ be polynomials with $Q$ not identically $0$. Then $Q$ has finitely many zeros (at most $\deg Q$ of them, by the fundamental theorem of algebra), so $\Omega = \{\, z : Q(z) \ne 0 \,\}$ is open. By part 3 of , the rational function $P/Q$ is holomorphic on $\Omega$ with
$$
\left(\frac{P}{Q}\right)' = \frac{P'Q - PQ'}{Q^2}.
$$
For example $f(z) = 1/z$ is holomorphic on $\mathbb{C}\setminus\{0\}$ with $f'(z) = -1/z^2$.
### 6.1. The exponential function
For $z = x+iy$ define
$$
e^{z} := e^{x}\bigl(\cos y + i \sin y\bigr),
$$
where $e^x$, $\cos y$, $\sin y$ on the right are the real functions.
Setting $y = 0$ recovers the real exponential, and setting $x = 0$ gives Euler's formula $e^{iy} = \cos y + i\sin y$ (). Defining the function by the power series $\sum_{n\ge0} z^n/n!$ yields the same function (see [Series and convergence tests](/mathematics/calculus/series-and-convergence)).
The function $e^z$ is holomorphic on $\mathbb{C}$, that is, entire, with $\dfrac{d}{dz}e^z = e^z$. Moreover, for all $z, w \in \mathbb{C}$ we have $e^{z+w} = e^z e^w$, $|e^z| = e^{\operatorname{Re} z} > 0$ (so $e^z$ never vanishes), and $e^{z + 2\pi i} = e^z$.
Here $u(x,y) = e^x\cos y$ and $v(x,y) = e^x \sin y$. The partial derivatives are
$$
u_x = e^x\cos y,\quad u_y = -e^x \sin y, \quad v_x = e^x\sin y, \quad v_y = e^x\cos y .
$$
These are continuous on all of $\mathbb{R}^2$, so $u, v$ are of class $C^1$. Also
$$
u_x = e^x\cos y = v_y, \qquad u_y = -e^x\sin y = -v_x ,
$$
so the Cauchy-Riemann equations hold in the whole plane. By , $e^z$ is entire and
$$
\frac{d}{dz}e^z = u_x + iv_x = e^x\cos y + i e^x \sin y = e^x(\cos y + i \sin y) = e^z .
$$
The addition formula follows from the definition together with the real exponential law and the addition formulas for sine and cosine. With $z = x+iy$ and $w = s+it$,
$$
e^z e^w = e^{x}e^{s}(\cos y + i\sin y)(\cos t + i \sin t)
= e^{x+s}\bigl[\cos(y+t) + i \sin(y+t)\bigr] = e^{z+w}
$$
(the middle step uses $(\cos y + i \sin y)(\cos t + i\sin t) = (\cos y\cos t - \sin y \sin t) + i(\sin y\cos t + \cos y \sin t)$).
For the modulus, $|e^z|^2 = e^{2x}(\cos^2 y + \sin^2 y) = e^{2x}$, so $|e^z| = e^x = e^{\operatorname{Re} z}$. This is a positive number, hence $e^z \ne 0$. Periodicity follows from the $2\pi$-periodicity of $\cos$ and $\sin$: $e^{z+2\pi i} = e^x(\cos(y+2\pi) + i\sin(y+2\pi)) = e^z$.
The greatest difference from the real exponential is **periodicity**. Since $e^z$ has period $2\pi i$, it is not injective. That fact will cause trouble when we define the logarithm.
### 6.2. Trigonometric functions
$$
\cos z := \frac{e^{iz} + e^{-iz}}{2}, \qquad \sin z := \frac{e^{iz} - e^{-iz}}{2i} .
$$
For real $z$, Euler's formula shows that these agree with the real $\cos$ and $\sin$.
The functions $\cos z$ and $\sin z$ are entire, with
$$
\frac{d}{dz}\cos z = -\sin z, \qquad \frac{d}{dz}\sin z = \cos z ,
$$
and $\cos^2 z + \sin^2 z = 1$ for all $z \in \mathbb{C}$.
The map $z \mapsto iz$ is entire () and $e^w$ is entire (), so by part 4 of the function $e^{iz}$ is entire with derivative $e^{iz}\cdot i = ie^{iz}$. Likewise $e^{-iz}$ is entire with derivative $-i e^{-iz}$. By part 1 of (linearity), $\cos z$ and $\sin z$ are entire as well.
Now the derivatives. Multiplying $\sin z = \dfrac{e^{iz}-e^{-iz}}{2i}$ by $i$ gives $\dfrac{e^{iz}-e^{-iz}}{2} = i \sin z$, which we record. Then
$$
\frac{d}{dz}\cos z = \frac{i e^{iz} + (-i) e^{-iz}}{2} = i\cdot\frac{e^{iz}-e^{-iz}}{2} = i \cdot i \sin z = -\sin z ,
$$
and similarly
$$
\frac{d}{dz}\sin z = \frac{i e^{iz} - (-i)e^{-iz}}{2i} = \frac{i\bigl(e^{iz}+e^{-iz}\bigr)}{2i} = \frac{e^{iz}+e^{-iz}}{2} = \cos z .
$$
Finally we prove $\cos^2 z + \sin^2 z = 1$. Put $A = e^{iz}$ and $B = e^{-iz}$; by the addition formula in , $AB = e^{iz - iz} = e^0 = 1$. Using $(2i)^2 = -4$,
$$
\cos^2 z + \sin^2 z = \frac{(A+B)^2}{4} + \frac{(A-B)^2}{-4}
= \frac{(A^2 + 2AB + B^2) - (A^2 - 2AB + B^2)}{4} = \frac{4AB}{4} = AB = 1 .
$$
On the real axis $|\sin x| \le 1$, but on the imaginary axis the situation is entirely different. For $z = iy$ with $y \in \mathbb{R}$,
$$
\sin(iy) = \frac{e^{i(iy)} - e^{-i(iy)}}{2i} = \frac{e^{-y} - e^{y}}{2i} = i\,\frac{e^{y}-e^{-y}}{2} = i \sinh y
$$
(using $1/i = -i$). Hence $|\sin(iy)| = |\sinh y| = \dfrac{e^{|y|} - e^{-|y|}}{2} \to \infty$ as $|y| \to \infty$. For example, $y = 10$ gives $|\sin(10 i)| = \sinh 10 \approx 1.1013 \times 10^{4}$.
Granting Liouville's theorem, that a bounded entire function is constant (), unboundedness of $\sin z$ is inevitable, since $\sin z$ is a nonconstant entire function. Boundedness on the real axis was an illusion produced by looking only at a "thin" subset of the plane.
### 6.3. The logarithm: multivaluedness and the principal branch
Since $e^z$ has period $2\pi i$, it is not injective and its inverse is not single valued. Let us solve $e^{\zeta} = w$ for $w \ne 0$. Writing $\zeta = \xi + i\eta$, gives $|e^\zeta| = e^{\xi}$, so $\xi = \ln|w|$ (the real logarithm). Then $e^{i\eta} = w/|w|$ says that $\eta$ is an argument of $w$, determined up to an integer multiple of $2\pi$. That is,
$$
\{\, \zeta : e^{\zeta} = w \,\} = \{\, \ln|w| + i(\arg w + 2\pi n) : n \in \mathbb{Z} \,\} .
$$
Fixing this indeterminacy at one choice is what a branch is.
For $z \in \mathbb{C}\setminus\{0\}$, among the values of $\arg z$ (infinitely many reals differing by integer multiples of $2\pi$) exactly one lies in the interval $(-\pi, \pi]$; it is called the **principal value of the argument** and denoted $\operatorname{Arg} z$. Then
$$
\operatorname{Log} z := \ln|z| + i \operatorname{Arg} z
$$
is called the **principal branch** of the logarithm.
To discuss holomorphy of the principal branch we must restrict to a region on which $\operatorname{Arg}$ is continuous, since $\operatorname{Arg}$ jumps from $\pi$ to near $-\pi$ across the negative real axis. So we consider the **cut** region
$$
\Omega_0 := \mathbb{C}\setminus(-\infty, 0] = \{\, r e^{i\theta} : r > 0,\ -\pi < \theta < \pi \,\} .
$$
To establish holomorphy there, a polar version of the equations is convenient.
Let $U \subset (0,\infty)\times(-\pi,\pi)$ be open and consider a function $f = u+iv$ on the image $\Phi(U)$ under $\Phi(r,\theta) = (r\cos\theta,\, r\sin\theta)$. If $u, v$ are of class $C^1$ as functions of $(r,\theta)$, then $f$ is holomorphic on $\Phi(U)$ if and only if
$$
\frac{\partial u}{\partial r} = \frac{1}{r}\frac{\partial v}{\partial \theta}, \qquad
\frac{\partial v}{\partial r} = -\frac{1}{r}\frac{\partial u}{\partial \theta}
$$
hold on $U$. In that case, at $z = re^{i\theta}$,
$$
f'(z) = e^{-i\theta}\left( \frac{\partial u}{\partial r} + i \frac{\partial v}{\partial r}\right) .
$$
The map $\Phi$ is a $C^\infty$ bijection from $(0,\infty)\times(-\pi,\pi)$ onto $\Omega_0$ with Jacobian determinant $\det\begin{pmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{pmatrix} = r > 0$, so by the inverse function theorem $\Phi^{-1}$ is $C^\infty$ as well. Hence "$u,v$ of class $C^1$ in $(r,\theta)$" and "$u,v$ of class $C^1$ in $(x,y)$" are equivalent, and by and all we must show is the equivalence of the Cartesian equations with the polar ones.
The chain rule gives
$$
\begin{aligned}
u_r &= u_x \cos\theta + u_y \sin\theta, & \frac{1}{r}u_\theta &= -u_x\sin\theta + u_y\cos\theta,\\
v_r &= v_x \cos\theta + v_y \sin\theta, & \frac{1}{r}v_\theta &= -v_x\sin\theta + v_y\cos\theta ,
\end{aligned}
$$
since $x = r\cos\theta$ and $y = r\sin\theta$ give $x_r = \cos\theta$, $y_r = \sin\theta$, $x_\theta = -r\sin\theta$, $y_\theta = r\cos\theta$.
**(Cartesian $\Rightarrow$ polar.)** Substitute $u_x = v_y$ and $u_y = -v_x$:
$$
\frac{1}{r}v_\theta = -v_x\sin\theta + v_y\cos\theta = u_y\sin\theta + u_x\cos\theta = u_r ,
$$
$$
\frac{1}{r}u_\theta = -u_x\sin\theta + u_y\cos\theta = -v_y\sin\theta - v_x\cos\theta = -v_r .
$$
**(Polar $\Rightarrow$ Cartesian.)** The four formulas above say that the passage from $(u_x,u_y)$ to $(u_r, u_\theta/r)$ is given by the rotation matrix $\begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}$. A rotation matrix is orthogonal, so its inverse is its transpose, and
$$
u_x = u_r\cos\theta - \frac{u_\theta}{r}\sin\theta, \qquad u_y = u_r \sin\theta + \frac{u_\theta}{r}\cos\theta
$$
(and similarly for $v$). Substituting the polar equations $u_r = v_\theta/r$ and $u_\theta/r = -v_r$ gives
$$
u_x = \frac{v_\theta}{r}\cos\theta + v_r \sin\theta, \qquad
v_y = v_r\sin\theta + \frac{v_\theta}{r}\cos\theta ,
$$
which agree, so $u_x = v_y$. Also
$$
u_y = \frac{v_\theta}{r}\sin\theta - v_r\cos\theta, \qquad
v_x = v_r\cos\theta - \frac{v_\theta}{r}\sin\theta = -u_y .
$$
**(Formula for the derivative.)** When $f$ is holomorphic, fix $z = re^{i\theta}$ and approach radially with $h = \Delta r\, e^{i\theta}$ where $\Delta r$ is a nonzero real:
$$
f'(z) = \lim_{\Delta r \to 0}\frac{f\bigl((r+\Delta r)e^{i\theta}\bigr) - f(re^{i\theta})}{\Delta r\, e^{i\theta}}
= e^{-i\theta} \frac{\partial}{\partial r}\bigl(u + iv\bigr)
= e^{-i\theta}\left(u_r + i v_r\right) ,
$$
since the complex derivative has the same value along every approach, so this restricted limit also equals $f'(z)$.
The function $\operatorname{Log} z$ is holomorphic on $\Omega_0 = \mathbb{C}\setminus(-\infty,0]$, with
$$
\frac{d}{dz}\operatorname{Log} z = \frac{1}{z} \qquad (z \in \Omega_0),
$$
and $e^{\operatorname{Log} z} = z$ for $z \in \Omega_0$.
For $z = re^{i\theta}$ with $r>0$ and $-\pi < \theta < \pi$ we have $\operatorname{Arg} z = \theta$, so
$$
u(r,\theta) = \ln r, \qquad v(r,\theta) = \theta .
$$
These are $C^\infty$ on $(0,\infty)\times(-\pi,\pi)$, with partial derivatives
$$
u_r = \frac{1}{r},\quad u_\theta = 0, \quad v_r = 0, \quad v_\theta = 1 .
$$
We check the polar equations. The first reads $\dfrac{1}{r}v_\theta = \dfrac{1}{r}\cdot 1 = \dfrac{1}{r} = u_r$, and the second reads $-\dfrac{1}{r}u_\theta = -\dfrac{1}{r}\cdot 0 = 0 = v_r$. So by , $\operatorname{Log}$ is holomorphic on $\Omega_0$ with
$$
\frac{d}{dz}\operatorname{Log} z = e^{-i\theta}\left(\frac{1}{r} + i\cdot 0\right) = \frac{1}{re^{i\theta}} = \frac{1}{z}.
$$
Finally $e^{\operatorname{Log} z} = e^{\ln r + i\theta} = e^{\ln r}(\cos\theta + i \sin\theta) = r e^{i\theta} = z$ by .
There is some freedom in choosing the cut $(-\infty,0]$: a branch can be built in the same way on the complement of any ray emanating from the origin. But the cut itself cannot be dispensed with altogether: **there is no continuous logarithm on all of $\mathbb{C}\setminus\{0\}$**.
Here is the reason. Suppose a continuous function $L : \mathbb{C}\setminus\{0\} \to \mathbb{C}$ satisfied $e^{L(z)} = z$. Follow the unit circle $\gamma(s) = e^{is}$, $s \in [0, 2\pi]$, and consider $\varphi(s) := \operatorname{Im} L(\gamma(s))$. From $e^{L(\gamma(s))} = e^{is}$, the difference $\varphi(s) - s$ is an integer multiple of $2\pi$. Since $\varphi$ is continuous, $(\varphi(s)-s)/(2\pi)$ is a continuous integer-valued function, hence a constant $n$. But $\gamma(0) = \gamma(2\pi) = 1$ forces $\varphi(0) = \varphi(2\pi)$, contradicting $\varphi(2\pi) - \varphi(0) = 2\pi$.
This "shift by $2\pi i$ after one loop" is later quantified as the integral $\displaystyle\oint_{|z|=1}\frac{dz}{z} = 2\pi i$, and becomes the starting point of the theory of residues (see [Cauchy's integral theorem and integral formula](/mathematics/complex-analysis/cauchy-integral-theorem) and [The residue theorem and applications to definite integrals](/mathematics/complex-analysis/residue-theorem)).
General powers $z^\alpha$ with $\alpha \in \mathbb{C}$ can likewise be defined by the principal branch as $z^{\alpha} := e^{\alpha \operatorname{Log} z}$. By part 4 of and , this is holomorphic on $\Omega_0$ with
$$
\frac{d}{dz} z^{\alpha} = e^{\alpha \operatorname{Log} z}\cdot \frac{\alpha}{z} = \alpha\, \frac{z^{\alpha}}{z} = \alpha z^{\alpha - 1},
$$
where the last equality again uses the same principal branch.
---
## 7. Exercises
Find all points at which $f(x+iy) = x^2 + i y^2$ is complex differentiable and compute $f'$ there. Is $f$ holomorphic at any point?
Here $u = x^2$ and $v = y^2$, with $u_x = 2x$, $u_y = 0$, $v_x = 0$, $v_y = 2y$. These are continuous on the whole plane, so $u, v$ are of class $C^1$ and in particular totally differentiable at every point.
Now examine the Cauchy-Riemann equations. The second, $u_y = -v_x$, reads $0 = 0$ and always holds. The first, $u_x = v_y$, reads $2x = 2y$, that is, $y = x$.
Hence by (or ), $f$ is complex differentiable exactly at the points of the line $y = x$, where
$$
f'(x + ix) = u_x + i v_x = 2x + i\cdot 0 = 2x .
$$
At points with $y \ne x$, the contrapositive of shows $f$ is not complex differentiable.
As for holomorphy, the line $y = x$ has no interior points: every neighborhood of every point contains points off the line. So by there is no point at which $f$ is holomorphic. This is the standard illustration that "many points of complex differentiability" does not mean holomorphic.
Let $\Omega$ be a domain and $f$ a holomorphic function on $\Omega$. Show that if $|f|$ is constant on $\Omega$, then $f$ is constant.
Write $f = u+iv$ and $|f|^2 = u^2 + v^2 = c$ with $c \ge 0$ constant.
If $c = 0$, then $u^2 + v^2 = 0$ gives $u = v = 0$ at every point, so $f \equiv 0$ is constant. Assume from now on that $c > 0$.
By , $u, v$ are of class $C^1$ on $\Omega$, so we may differentiate $u^2 + v^2 = c$ with respect to $x$ and to $y$:
$$
2u u_x + 2v v_x = 0, \qquad 2u u_y + 2v v_y = 0,
$$
that is, $u u_x + v v_x = 0$ and $u u_y + v v_y = 0$. Substituting the Cauchy-Riemann equations $u_y = -v_x$, $v_y = u_x$ () into the second gives
$$
-u v_x + v u_x = 0 .
$$
Viewing this together with the first as a linear system in $(u_x, v_x)$, the coefficient matrix is
$$
\begin{pmatrix} u & v \\ v & -u \end{pmatrix}, \qquad
\det = -u^2 - v^2 = -c \ne 0 .
$$
Since the determinant is nonzero, the only solution is the trivial one, so $u_x = v_x = 0$ at every point of $\Omega$. By part 3 of ,
$$
f' = u_x + i v_x = 0
$$
on $\Omega$. As $\Omega$ is a domain, hence connected, shows that $f$ is constant.
Connectedness cannot be dropped. On $\Omega = D(0,1) \cup D(3,1)$, two disjoint disks, set $f = 1$ on the left disk and $f = -1$ on the right one. Then $f$ is holomorphic with $|f| \equiv 1$, yet it is not constant.
Verify that $u(x,y) = x^3 - 3xy^2$ is harmonic on $\mathbb{R}^2$, and find all real-valued functions $v$ for which $f = u + iv$ is entire. Then express the resulting $f$ as a formula in $z$ alone.
**Harmonicity.** From $u_x = 3x^2 - 3y^2$ we get $u_{xx} = 6x$, and from $u_y = -6xy$ we get $u_{yy} = -6x$. Hence $u_{xx} + u_{yy} = 6x - 6x = 0$ and $u$ is harmonic.
**Determining $v$.** For $f = u+iv$ to be holomorphic it suffices that $v$ be of class $C^1$ and satisfy the Cauchy-Riemann equations (). Integrating the first equation $v_y = u_x = 3x^2 - 3y^2$ in $y$ gives
$$
v(x,y) = 3x^2 y - y^3 + \varphi(x),
$$
the constant of integration being a function $\varphi$ of $x$. Substituting into the second equation $v_x = -u_y = 6xy$ gives
$$
v_x = 6xy + \varphi'(x) = 6xy \quad\Longrightarrow\quad \varphi'(x) = 0 .
$$
Since $\varphi$ is differentiable on $\mathbb{R}$ with vanishing derivative, the mean value theorem makes it a constant $C \in \mathbb{R}$. Therefore
$$
v(x,y) = 3x^2 y - y^3 + C .
$$
This $v$ is a polynomial, hence $C^\infty$, and it does satisfy the equations, so $f$ is entire by . Conversely, if $f$ is entire the equations are necessary (), so there is no other $v$.
**Formula in $z$.** Expanding $(x+iy)^3$,
$$
z^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = (x^3 - 3xy^2) + i\,(3x^2y - y^3),
$$
so $f(z) = u + iv = z^3 + iC$.
Answer the following about the principal branch $\operatorname{Log}$.
1. Let $x_0 < 0$ be real. Compute the limits of $\operatorname{Log}(x_0 + i\varepsilon)$ and $\operatorname{Log}(x_0 - i\varepsilon)$ as $\varepsilon \downarrow 0$, and confirm that $\operatorname{Log}$ admits no continuous extension across the negative real axis.
2. Give an example of $z_1, z_2$ for which $\operatorname{Log}(z_1 z_2) = \operatorname{Log} z_1 + \operatorname{Log} z_2$ fails, and compute the difference of the two sides.
**1.** Let $x_0 < 0$ and put $r_0 = |x_0| = -x_0$. For small $\varepsilon > 0$, the point $x_0 + i\varepsilon$ lies in the second quadrant, with $|x_0 + i\varepsilon| \to r_0$ and $\operatorname{Arg}(x_0+i\varepsilon) \to \pi$ (the argument approaches $\pi$ from slightly below). Hence
$$
\lim_{\varepsilon\downarrow 0}\operatorname{Log}(x_0+i\varepsilon) = \ln r_0 + i\pi .
$$
On the other hand $x_0 - i\varepsilon$ lies in the third quadrant with $\operatorname{Arg}(x_0 - i\varepsilon) \to -\pi$, so
$$
\lim_{\varepsilon\downarrow 0}\operatorname{Log}(x_0-i\varepsilon) = \ln r_0 - i\pi .
$$
The two limits differ by $2\pi i \ne 0$, so the limit at $x_0$ does not exist and $\operatorname{Log}$ cannot be made continuous at $x_0$. The principal value of does assign the value $\ln r_0 + i\pi$ at $x_0$ itself, but that merely agrees with the limit from above and does not restore continuity.
**2.** Take $z_1 = z_2 = e^{2\pi i/3}$. Since $|z_1| = 1$ and $\operatorname{Arg} z_1 = 2\pi/3 \in (-\pi,\pi]$, we have $\operatorname{Log} z_1 = \operatorname{Log} z_2 = i\,2\pi/3$, so
$$
\operatorname{Log} z_1 + \operatorname{Log} z_2 = \frac{4\pi i}{3}.
$$
On the other hand $z_1 z_2 = e^{4\pi i /3} = e^{-2\pi i/3}$ (because $4\pi/3 - 2\pi = -2\pi/3$), and $-2\pi/3 \in (-\pi,\pi]$, so $\operatorname{Arg}(z_1z_2) = -2\pi/3$ and
$$
\operatorname{Log}(z_1 z_2) = -\frac{2\pi i}{3}.
$$
The difference is
$$
\bigl(\operatorname{Log} z_1 + \operatorname{Log} z_2\bigr) - \operatorname{Log}(z_1z_2) = \frac{4\pi i}{3} + \frac{2\pi i}{3} = 2\pi i .
$$
The discrepancy of $2\pi i$ arose because the sum of the arguments left the range $(-\pi,\pi]$ of the principal value. In general one has $\operatorname{Log}(z_1z_2) = \operatorname{Log} z_1 + \operatorname{Log} z_2 + 2\pi i n$ with $n \in \{-1,0,1\}$.
---
## References
- L. V. Ahlfors, *Complex Analysis*, 3rd ed., McGraw-Hill, 1979 — Chapter 2, "Complex Functions", covers complex differentiability, the Cauchy-Riemann equations and the elementary functions.
- E. M. Stein and R. Shakarchi, *Complex Analysis*, Princeton University Press, 2003 — Chapter 1 follows the same line as this article: complex differentiation, Cauchy-Riemann, Wirtinger derivatives.
- Reiji Takahashi, *Shinpan Fukuso Kaiseki*, University of Tokyo Press, 1990 (in Japanese) — a standard Japanese textbook with a careful treatment of the definition and basic properties of holomorphic functions.
- Michio Jimbo, *Fukuso Kansu Nyumon*, Iwanami Shoten (Introduction to Modern Mathematics), 2003 (in Japanese) — Chapter 1; concrete handling of branches of the exponential and logarithmic functions.
- R. Narasimhan and Y. Nievergelt, *Complex Analysis in One Variable*, 2nd ed., Birkhäuser, 2001 — includes a proof of the Looman-Menchoff theorem.
- W. Rudin, *Real and Complex Analysis*, 3rd ed., McGraw-Hill, 1987 — Chapter 10; the logical order by which $C^\infty$ smoothness is derived from the definition of holomorphy is made explicit.
---
## Appendix: How far can the hypotheses be weakened?
**The question.** In the main text we showed that "$u,v$ of class $C^1$ plus the equations" yields holomorphy (), and that the pointwise necessary and sufficient condition is "totally differentiable plus the equations" (). On the other hand, showed that mere existence of the partial derivatives is not enough. What conditions lie in between? The question was studied intensively in the first half of the twentieth century, and several definitive answers are known.
**Goursat's theorem.** Historically the first improvement was the removal of continuity of $f'$ from the hypotheses of Cauchy's integral theorem. Cauchy's own proof went through Green's theorem and thus required $u,v \in C^1$; around 1900 Goursat showed that the integral over a triangle vanishes assuming only holomorphy, that is, pointwise complex differentiability. Thanks to this improvement, the chain "holomorphic $\Rightarrow$ infinitely differentiable $\Rightarrow$ $u,v \in C^\infty$" runs without circularity. The implication 1 $\Rightarrow$ 2 of is a corollary of that result. The details are treated in [Cauchy's integral theorem and integral formula](/mathematics/complex-analysis/cauchy-integral-theorem).
**The Looman-Menchoff theorem.** A still stronger result is known: let $\Omega$ be open and $f : \Omega \to \mathbb{C}$ **continuous**, and suppose the four partial derivatives of $u = \operatorname{Re} f$ and $v = \operatorname{Im} f$ exist at **every point** of $\Omega$ and satisfy the Cauchy-Riemann equations; then $f$ is holomorphic on $\Omega$. Neither total differentiability nor continuity of the partial derivatives is needed. The continuity hypothesis, however, cannot be dropped: the function $e^{-z^{-4}}$ of is holomorphic away from the origin and has partial derivatives at the origin satisfying the equations, yet it is not holomorphic there because it is not continuous. The proof is real-analytic, using tools such as the Baire category theorem, and lies off the standard path of complex analysis (see the book of Narasimhan and Nievergelt).
**A practical moral.** These refinements are mainly of technical interest; for ordinary applications the $C^1$ test of is enough. If the $u, v$ at hand are built from elementary functions by arithmetic operations and composition, then $C^1$ regularity is automatic, and the only thing left to verify is the two equations.