Lecture 5: Complex Differentiation

Complex differentiation

With the concept of limit and continuity, we are ready to introduce differentiation on the complex plane. Complex differentiation is defined as follows:

Definition 1 (Complex Derivative). We say that $f(z):\mathbb C\to \mathbb C$ is complex differentiable at $z_0$ (or more precisely in a neighborhood of $z_0$ ) if the limit $\lim_{z\to z_0}\dfrac{f(z)-f(z_0)}{z-z_0}$ exists. In this case, we denote the limit by $f'(z_0)$ or $\dfrac{df}{dz}(z_0)$ and call it the the complex derivative of $f$ at $z_0$ .

Clearly, if we introduce $\delta z=z-z_0$ , we have the following equivalent form of the definition of complex derivative: $f'(z_0) := \lim_{\delta z\to 0}\dfrac{f(z_0+\delta z)-f(z_0)}{\delta z}.$

It is also immediately clear that for a constant function $f(z)=c$ , $f'(z)=0$ .

Definition 2 (Complex Analytic/Holomorphic Function). A complex function $f:\mathbb C\to\mathbb C$ is said to be analytic at $z$ if it is complex differentiable at $z$ . A function $f$ is said to be complex analytic or holomorphic in a domain $\Omega\subseteq\mathbb C$ if it is complex differentiable in a neighborhood of every point in $\Omega$ . A function is said to be entire if it is holomorphic on the whole complex plane $\mathbb C$ .

Definition 3 (Singular Point). A singular point of $f(z)$ is a point $z_0\in Dom(f)$ where $f'(z_0)$ does not exist.

A singular point of a function $f$ is often loosely called a singularity of $f$ .

In the same manner as we learned in functions of real variables, if a complex function is differentiable at $z$ , it is continuous at $z$ .

Theorem 4 (Differentiability Implies Continuity). If $f(z):\mathbb C\to \mathbb C$ is differentiable at $z_0$ , then $f(z)$ is continuous at $z_0$ .

Proof. This follows from the fact that $\displaystyle\lim_{z \to z_0} f(z) = \displaystyle\lim_{z\to z_0}[\dfrac{f(z)-f(z_0)}{z-z_0} (z-z_0) + f(z_0)]=f'(z_0)0+f(z_0) = f(z_0)$ . ◻

Most of the consequences of differentiability are quite different in the real and complex cases. We will see this in the next a few lectures. However, the simplest algebraic rules of differentiation are the same in the real and complex cases. We now briefly review those algebraic rules.

Theorem 5 (Properties of Complex Derivatives). Let $f(z): \mathbb C\to \mathbb C$ and $g(z): \mathbb C\to \mathbb C$ be differentiable at $z$ with derivatives $f'(z)$ and $g'(z)$ respectively. Then $(f+g)'(z) = f'(z)+g'(z)$ $(fg)'(z) = f'(z)g(z)+f(z)g'(z)$ $(f/g)'(z) = (f'(z)g(z)-f(z)g'(z))/g^2(z),\ \ \ g(z) \neq 0$ $[f(g)(z)]' = f'(g(z)) g'(z), \ \ \ \mbox{if}\ f'(g(z))\ \mbox{exists}.$

Proof. The results follows directly from the definition and the properties of limit. We only prove the second and the fourth property. $\begin{gathered} (fg)'(z)=\displaystyle\lim_{\delta z\to 0}\dfrac{f(z+\delta z)g(z+\delta z)-f(z)g(z)}{\delta z}\\ =\displaystyle\lim_{\delta z\to 0}\dfrac{f(z+\delta z)g(z+\delta z)-f(z+\delta z)g(z) + f(z+\delta z)g(z)-f(z)g(z)}{\delta z} \\ =\displaystyle\lim_{\delta z\to 0}\dfrac{f(z+\delta z)(g(z+\delta z)-g(z))}{\delta z} + \dfrac{(f(z+\delta z)-f(z))g(z)}{\delta z}\\ =f(z)g'(z)+f'(z)g(z). \end{gathered}$ $\begin{gathered} (f(g))'(z)=\displaystyle\lim_{\delta z\to 0}\dfrac{f(g(z+\delta z))-f(g(z))}{\delta z}=\displaystyle\lim_{\delta z\to 0}\dfrac{f(g(z+\delta z))-f(g(z))}{g(z+\delta z)-g(z)}\dfrac{g(z+\delta z)-g(z)}{\delta z} \\ =\displaystyle\lim_{\delta z\to 0}\dfrac{f(\frac{g(z+\delta z)-g(z)}{\delta z} \delta z+g(z))-f(g(z))}{\frac{g(z+\delta z)-g(z)}{\delta z} \delta z} \cdot \dfrac{g(z+\delta z)-g(z)}{\delta z}\\ =f'((g(z)) g'(z). \end{gathered}$ ◻

Exercise 6. Prove the results in Theorem Theorem 5.

Theorem 7 (Derivatives of Elementary Complex Functions). Let $n\ge 1$ be an integer. Then $(a)\ \dfrac{d}{dz} z^n = n z^{n-1},$ $(b)\ \dfrac{d}{dz} e^z = e^z,$ $(c)\ \dfrac{d}{dz} \sin z = \cos z,$ $(d)\ \dfrac{d}{dz} \cos z = -\sin z .$

Proof. We prove the first two results. (a) Using the binomial formula $(z+\delta z)^n=\sum_{j=1}^n \dfrac{n!}{j!(n-j)!}z^j(\delta z)^{n-j}$ , we see immediately that $\dfrac{d}{dz} z^n = \displaystyle\lim_{\delta \to 0}\dfrac{(z+\delta z)^n-z^n}{\delta z}=\displaystyle\lim_{\delta z\to 0} \sum_{j=1}^{n-1} \dfrac{n!}{j!(n-j)!}z^j(\delta z)^{n-j-1}.$ The only term survives in the limit is the $j=n-1$ term which gives $nz^{n-1}$ .
(b) We first prove that $\displaystyle\lim_{\delta z\to 0} \dfrac{e^{\delta z}-1}{\delta z} =1$ . This follows from the following calculation: $\displaystyle\lim_{\delta z\to 0} e^{\delta z} =\displaystyle\lim_{\delta z\to 0} e^{\delta x +i \delta y} =\displaystyle\lim_{\delta z\to 0} e^{\delta x} e^{i \delta y} = \displaystyle\lim_{\delta x+ i\delta y\to 0} e^{\delta x} (\cos \delta y + i\sin \delta y) =1$ . We now have $\displaystyle\lim_{\delta z\to 0}\dfrac{e^{z+\delta z}-e^z}{\delta z}=\displaystyle\lim_{\delta z\to 0}e^z \dfrac{e^{\delta z}-1}{\delta z} = e^z$ . ◻

Exercise 8. Prove (c) and (d) of Theorem Theorem 7.

Remark 9. The fact that when $f(z)$ is continuous, $\overline{f(z)}$ is also continuous could easily lead us to believe that when $f(z)$ is differentiable, $\overline{f(z)}$ is also differentiable. However, this is Not true. For instance, $f(z)=z$ is differentiable. To see why, we observe that $\lim_{\delta z\to 0}\dfrac{\overline{z+\delta z}-\bar z}{\delta z} = \lim_{\delta z\to 0}\dfrac{\overline{\delta z}}{\delta z} = \lim_{\delta z\to 0} e^{-i2\theta}$ with $\theta$ the argument of $\delta z$ . If we take $\delta z\to 0$ along the positive $x$ -axis ( $\theta=0$ ), we have that the limit is $1$ . If we take $\delta z \to 0$ along the positive $y$ -axis ( $\theta=\pi/2$ ), we have that the limit is $-1$ . Therefore, the taking limit from different paths yield different results. This means that the limit does not exist!

The Cauchy-Riemann equations

Complex functions that are differentiable have some special properties. For instance, if a function is differentiable at $z$ , it is continuous there ([the converse is NOT true]). We now look at some other property of differentiable complex functions.

One of the most important results about differentiability is following theorem.

Theorem 10 (Cauchy-Riemann Conditions). The complex function $f(z) = u(x,y)+iv(x,y)$ is differentiable at $z=x+iy$ of a region if and only if the partial derivatives $u_x$ , $u_y$ , $v_x$ and $v_y$ are continuous and satisfy the following Cauchy-Riemann equations: $\label{EQ:CR} u_x = v_y, \ \ \ v_x = -u_y \,.$

Proof. (i) If $f$ is differentiable at $z$ , then $f'(z)=\displaystyle\lim_{\delta z\to 0}\dfrac{f(z+\delta z)-f(z)}{\delta z}$ . Take $\delta z=\delta x$ , we have that $\begin{gathered} f'(z)=\displaystyle\lim_{\delta x \to 0}\dfrac{f(z+\delta x)-f(z)}{\delta x} \\ =\lim_{\delta x\to 0}\dfrac{(u(x+\delta x, y)-u(y))+i(v(x+\delta x, y)-v(x,y))}{\delta x} = u_x(x,y) + i v_x(x,y) \end{gathered}$ Take $\delta z=i\delta y$ , we have that $\begin{gathered} f'(z)=\displaystyle\lim_{\delta y \to 0}\dfrac{f(z+i \delta y)-f(z)}{i \delta y} \\ =\lim_{\delta y\to 0}\dfrac{(u(x, y+\delta y)-u(y))+i(v(x, y+\delta y)-v(x,y))}{\delta y} = -iu_y(x,y) + v_y(x,y) \end{gathered}$ The two ways of calculation should result in the same result. Therefore, we have [EQ:CR].

The reason why the partial derivatives have to be continuous is hard to prove right now. We postpone that part to a future lecture.

(ii) If the partial derivatives are continuous, then by Taylor’s Theorem, we know that $u(x+\delta x, y+ \delta y)=u(x, y)+ u_x \delta x+ u_y \delta y + p(x, y) \sqrt{(\delta x)^2+(\delta y)^2}$ with $p\to 0$ when $(\delta x, \delta y)\to 0$ . Similarly, $v(x+\delta x, y+ \delta y)=v(x, y)+ v_x \delta x+ v_y \delta y + q(x, y) \sqrt{(\delta x)^2+(\delta y)^2}$ with $q\to 0$ when $(\delta x, \delta y)\to 0$ . Therefore, after some algebraic calculations, we have $\dfrac{f(z+\delta z)-f(z)}{\delta z}=(u_x\dfrac{\delta x}{\delta z}+u_y\dfrac{\delta y}{\delta z})+i(v_x\dfrac{\delta x}{\delta z}+v_y\dfrac{\delta y}{\delta z})+(p+iq)\dfrac{|\delta z|}{\delta z}.$ If the Cauchy-Riemann equations are satisfied, then this simplifies to $\dfrac{f(z+\delta z)-f(z)}{\delta z}=(u_x+iv_x) \dfrac{\delta x+i\delta y}{\delta z}+(p+iq)\dfrac{|\delta z|}{\delta z}.$ The first term on the right is $f'(z)$ and the second term on the right vanishes when taking the limit $\delta z\to 0$ . Therefore, $f(z)$ is differentiable. ◻

Remark 11. An important observation from the proof of the above theorem is that $f'(z) = \dfrac{\partial f}{\partial x} \ \ \ \mbox{and}\ \ \ f'(z) = -i\dfrac{\partial f}{\partial y}$ The Cauchy-Riemann equation can therefore be written as $\dfrac{\partial f}{\partial x}= -i\dfrac{\partial f}{\partial y} .$

Therefore, to check whether or not a complex function $f(z): \mathbb C\to \mathbb C$ is analytic at a point, we need to check whether or not the Cauchy-Riemann equation is satisfied in a neighborhood of that point.

Example 12. The function $f(z)=e^z$ is analytic everywhere. To see that, we observe that $f(z)=e^{x+iy}=e^xe^{iy}=e^x\cos y + i e^x\sin y$ . Therefore $u(x,y)=e^x \cos y$ and $v(x,y)=e^x\sin y$ . It is easy to verify that $u_x=e^x\cos y=v_y$ , and $u_y=-e^x\sin y= -v_x$ hold everywhere on $\mathbb C$ .

Example 13. The function $f(z)=z^2$ is analytic everywhere. To see that, we observe that $f(z)=z^2=(x^2-y^2)+i2xy$ . Therefore $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$ . It is easy to verify that $u_x=2x=v_y$ , and $u_y=-2 y= -v_x$ hold everywhere on $\mathbb C$ .

Example 14. The function $f(z)=|z|$ is nowhere analytic.

Proof. (By Contradiction) Let $u(x,y)=\sqrt{x^2+y^2}$ and $v(x,y)=0$ . Then $f(z)=|z|=u(x,y)+iv(x,y)$ . We then have $u_x(x,y)=\dfrac{2x}{\sqrt{x^2+y^2}}, \quad v_y=0, \quad u_y(x,y)=\dfrac{2y}{\sqrt{x^2+y^2}}, \quad -v_x=0.$ Assume that $f$ is differentiable at some $z$ . Then the Cauchy-Riemann conditions require that $u_x=u_y=0$ . This means that $\dfrac{2x}{\sqrt{x^2+y^2}}=\dfrac{2y}{\sqrt{x^2+y^2}}=0$ which is impossible. Therefore $f$ is nowhere differentiable. ◻

Consequences of analyticity

Being analytic has huge consequences. Let us define $n$ -th $(n\ge 2)$ order derivatives of $f(z):\mathbb C\to \mathbb C$ as $f^{(n)}(z) \equiv \displaystyle\lim_{\delta z\to 0}\dfrac{f^{(n-1)}(z+\delta z)-f^{(n-1)}(z)}{\delta z}$ . Then we can prove the following result, which we will do in a few lectures.

Theorem 15 (Analytic Functions Are Differentiable Infinitely Many Times). Let $f(z)\equiv u(x,y)+iv(x,y)$ be analytic in region $\Omega$ . Then $f^{(k)}(z)$ exists in $\Omega$ for any $k\ge 1$ . Moreover, $u(x,y)$ and $v(x,y)$ have continuous partial derivatives of any order.

The Laplace Equation.

By the theorem, we can taking partial derivatives of $u$ and $v$ of any order. Let’s differentiate the Cauchy-Riemann equations to obtain $u_{xx}=v_{yx}, \ \ \ v_{xy} = -u_{yy}$ This immediately gives that $\Delta u := u_{xx}+u_{yy} =0.$ This equation for $u$ is called the Laplace equation.

In the same manner, we can show that $v$ satisfy the Laplace equation also: $\Delta v : = v_{xx}+v_{yy} =0.$ Therefore, if a function is analytic, then the real and imaginary parts of the function both satisfy the Laplace equation.

Notation 1. The Laplace operator $\Delta$ is sometimes denoted by $\nabla^2$ .

Definition 16 (Harmonic Functions). A function $f:\mathbb R^2\to\mathbb R$ that satisfies the Laplace equation is called is a harmonic function. The real and imaginary parts of an analytic function is called the harmonic conjugate of each other.

It is clear that two functions that are harmonic conjugate of each other, that is the real and imaginary parts an analytical function, are not independent since they are tied together by the Cauchy-Riemann equations. Therefore it is possible to find one given the other. Here is an example.

Example 17. Let $f(z)=u(x,y)+iv(x,y)$ be an analytical function. Assume that $u(x,y)=x(1-y)$ , find $f(z)$ .

To solve this problem, we note that . Therefore, by the Cauchy-Riemann equation, we have $v_y=u_x=1-y, \ \ \ v_x = -u_y = x.$ Therefore, we can integrate these equations to obtain $v(x,y) = y-y^2/2+p(x), \ \ \ v(x,y) = x^2/2 + q(y)$ Therefore $p(x)=x^2/2+c$ and $q(y)=y-y^2/2+c$ with $c$ an arbitrary constant.

Therefore $f(z) = x(1-y) + i(x^2/2+y-y^2/2+c)$ .

Example 18. Let $f(z)=u(x,y)+iv(x,y)$ be an analytical function. Assume that $v(x,y)=x-y$ , find $f(z)$ .

To solve this problem, we note that . Therefore, by the Cauchy-Riemann equation, we have $u_x=v_y=-1, \ \ \ u_y=-v_x= -1.$ Therefore, we can integrate these equations to obtain $u(x,y) = -x+p(y), \ \ \ u(x,y) = -y + q(x)$ Therefore $p(y)=-y+c$ and $q(x)=-x+c$ with $c$ an arbitrary constant.

Therefore $f(z) = (x-y) + i(-x-y+c)$ .

Remark 19. Motivated by the fact that $x=(z+\bar z)/2$ , $y=(z-\bar z)/(2i)$ , which, if $z$ and $\bar z$ were independent variables, would give $\partial x/\partial z=1/2$ and $\partial y/\partial z=-i/2$ , it is convenient to introduce the notations: $\frac{\partial f}{\partial z} \equiv \frac12\left(\frac{\partial f}{\partial x}-i\frac{\partial f}{\partial y}\right), \qquad \frac{\partial f}{\partial \bar z} \equiv \frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}\right).$ In terms of these notations, the Cauchy–Riemann equations are exactly equivalent to $\frac{\partial f}{\partial \bar z}=0,$ which is also equivalent to $\frac{\partial f}{\partial z}=\frac{\partial f}{\partial x}.$

Moreover, it is easy to verify that $\label{EQ:Complex Derivative2} \frac{\partial}{\partial z}\frac{\partial}{\partial \bar z} =\frac{\partial}{\partial\bar z}\frac{\partial}{\partial z} =\frac14\Delta.$ This shows that any analytic function is harmonic (equivalently, its real and imaginary parts are harmonic). It also shows that the conjugate of an analytic function, while not analytic, is harmonic.

Exercise 20. Verify [EQ:Complex Derivative2].

Guided review and additional examples

Learning goals

Test a difference quotient along arbitrary complex directions.
Use the Cauchy–Riemann equations as a practical differentiability test.
Distinguish differentiability at one point from analyticity on a neighborhood.
Compute derivatives using complex analogues of familiar rules.

Concept connection: the derivative must ignore direction

The quotient $[f(z+h)-f(z)]/h$ must approach one value as the complex increment $h$ approaches zero from every direction. The Cauchy–Riemann equations express the compatibility forced by that requirement. They are necessary at a differentiability point; with continuous first partial derivatives in a neighborhood, they are also sufficient there.

Worked example 1: verify $f(z)=z^2$

Write $f=u+iv$ with $u=x^2-y^2$ and $v=2xy$ . Then

$u_x=2x=v_y,\qquad u_y=-2y=-v_x.$

The equations hold everywhere, so $f$ is entire. Its derivative is

$f'(z)=u_x+i v_x=2x+2iy=2z.$

Worked example 2: $f(z)=|z|^2$ at the origin

Here $u=x^2+y^2$ and $v=0$ . The Cauchy–Riemann equations hold only at $(0,0)$ . Directly,

$\frac{f(h)-f(0)}{h}=\frac{|h|^2}{h}=\overline h\longrightarrow0.$

So $f$ is complex differentiable at $0$ with $f'(0)=0$ , but it is not analytic at $0$ because it is not differentiable throughout any neighborhood of $0$ .

Check your understanding

Why does checking only horizontal and vertical approaches to a difference quotient not by itself prove differentiability?

Show the answer

Those checks examine only two of infinitely many directions and paths. Agreement is necessary, but a different slanted or curved approach may still produce another limit. A complete proof needs a direction-independent estimate or an applicable theorem such as the Cauchy–Riemann sufficiency result.