Complex Differentiation: Definition, Solved Problems

In the post, we will learn about complex differentiation where we study the derivative of functions of a complex variable along with some solved problems.

ℂ := The set of complex numbers.

Complex Differentiation: Definition

Let D ⊆ ℂ be an open set and let f: D→ℂ be a complex function. The function f is said to be differentiable (or complex differentiable) at z₀ ∈ D if the limit

lim_h→0 $\dfrac{f(z_0+h)-f(z_0)}{h}$

exists. In this case, the value of the limit is called the derivative of f at z₀, denoted by f$’$(z₀).

Note:

1. The above limit is same as $\lim\limits_{z \to z_0}\dfrac{f(z)-f(z_0)}{z -z_0}$. So its value is equal to f$’$(z₀).
2. If the function f is differentiable at every point in the complex plane, then f is called an entire function.

Example

For a natural number n, the function f(z) = zⁿ is differentiable at every point z₀ ∈ ℂ.

Proof:

Let z₀ ∈ ℂ be fixed.

Now,

$\lim\limits_{z \to z_0}\dfrac{f(z)-f(z_0)}{z -z_0}$

= $\lim\limits_{z \to z_0}\dfrac{z^n-z_0^n}{z -z_0}$

= $\lim\limits_{z \to z_0}$ [z^n-1 + z^n-2 z₀ + … + z₀^n-1]

= $nz_0^{n-1}$.

So the derivative of f(z)=zⁿ is equal to $nz_0^{n-1}$, that is, $f'(z_0)=nz_0^{n-1}$. As z₀ is an arbitrary point, we conclude that zⁿ is differentiable at every point in the complex plane. This makes the function f(z) = zⁿ entire.

Remark: f(z) = zⁿ is an example of an entire function.

Properties

If two functions f and g are differentiable at z₀, then their sum, difference, product and quotient (denominator never vanishes), scalar multiplication are also differentiable. In other words,

• $(f+g)’ = f’+g’$
• $(f-g)’ = f’-g’$
• $(fg)’ = f’g+fg’$
• $\left(\dfrac{f}{g} \right)’$ = $\dfrac{gf’-fg’}{g^2}$
• $(cf)’ = cf’$, c is a constant.

Theorems

Theorem 1: If a function f(z) is differentiable z0, then it is continuous at z0.

Proof:

Assume that f(z) is differentiable at z₀. So the limit $\lim\limits_{z \to z_0}\dfrac{f(z)-f(z_0)}{z -z_0}$ exists, and equals to $f'(z_0)$. To show f is continuous at z₀, we need to show the follwoing:

$\lim\limits_{z \to z_0}f(z)=f(z_0)$.

Now,

$\lim\limits_{z \to z_0}[f(z)-f(z_0)]$

= $\lim\limits_{z \to z_0}$ $\left[ \dfrac{f(z)-f(z_0)}{z -z_0}\times (z-z_0) \right]$

= $\lim\limits_{z \to z_0}\dfrac{f(z)-f(z_0)}{z -z_0}$ $\times \lim\limits_{z \to z_0} (z-z_0)$

= $f'(z_0) \times 0$

= 0.

Therefore, $\lim\limits_{z \to z_0}[f(z)-f(z_0)]$ = 0 which implies that $\lim\limits_{z \to z_0}f(z)=f(z_0)$. This proves that f(z) is continuous at z=z₀.

Theorem 2: The derivative of a real-valued function of a complex variable either zero or the derivative does not exist.

Proof:

Let f(z) be a real-valued function of a complex variable, that is, f: ℂ→ℝ.

Let a∈ℂ be arbitrary. Then

$f'(a)= \lim\limits_{h \to 0} \dfrac{f(a+h)-f(a)}{h}$.

We evaluate this limit along the following two paths.

• Take h→0 along the real axis. Then f$’$(a) is a real number.
• Take h→0 along the imaginary axis. Then f$’$(a) is a purely imaginary number.

Thus, we conclude that f$’$(a) is either zero or it does not exist.

Solved Problems

Q1: Show that f(z) = |z| is not differentiable at the origin z=0.

Answer:

Let f(z) = |z|. To show it is not differentiable at z=0, we need to verify the limit

lim_h→0 $\dfrac{f(h)-f(0)}{h}$

does not exist. Note that

lim_h→0 $\dfrac{f(h)-f(0)}{h}$ = lim_h→0 $\dfrac{|h|}{h}$.

Now evaluate this limit along the following two paths.

1. Let h = h₁+i⋅0 with 0<h₁. See h₁→0 as h→0. Along this path, the above limit lim_h→0 $\dfrac{f(h)-f(0)}{h}$ is equal to 1.
2. Let h = h₁+i⋅0 with 0>h₁. Then h₁→0 as h→0. In this case, lim_h→0 $\dfrac{f(h)-f(0)}{h}$ = -1.

As we have obtained two limits for two different paths, the above limit does not exist. In other words, f(z) = |z| (mod z) is not differentiable at z=0.

Q2: Discuss the differentiability of f(z) = |z|2.

Answer:

Given f(z) = |z|².

First, we discuss the differentiability of f(z) = |z|² at z₀=0. Note that

$\lim\limits_{h \to 0}\dfrac{f(h)-f(0)}{h}$ = $\lim\limits_{h \to 0}\dfrac{|h|^2}{h}$ = $\lim\limits_{h \to 0}\dfrac{h\overline{h}}{h}$ = $\lim\limits_{h \to 0}\overline{h}$ = 0.

Thus, the function f(z) = |z|² is differentiable at z₀=0.

Now, consider z₀ ≠ 0. Then

lim_h→0 $\dfrac{f(z_0+h)-f(z_0)}{h}$

= lim_h→0 $\dfrac{|z_0+h|^2-|z_0|^2}{h}$

= lim_h→0 $\dfrac{(z_0+h) (\overline{z_0+h})-z_0 \overline{z_0}}{h}$

= lim_h→0 $\dfrac{(z_0+h)(\overline{z_0}+\overline{h})-z_0 \overline{z_0}}{h}$

= lim_h→0 $\dfrac{\overline{z_0}h +z_0\overline{h} +h\overline{h}}{h}$

As $\dfrac{\overline{h}}{h}$ tends to different values when h→0, we conclude that the function f(z) = |z|² is not differentiable at non-zero points z₀.

Homepage of Complex Analysis

Practice Questions

1. For a fixed a∈ℂ, show that the function f(z) = |z-a|² is differentiable at z=a only.
2. Show that f(z) = z Re(z) is differentiable at z=0 only and nowhere else.
3. Prove that f(z) = Re(z) and g(z) = Im(z) are nowhere differentiable.
4. Show $\overline{z}$, Arg(z) are nowhere differentiable.

FAQs