3.2 Complex Derivatives

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We find derivatives of complex functions

We begin with the derivative of a complex function at $z = z_0$ .

If $f(z)$ is defined in $D(z_0, r)$ for some $r>0$ , then we say that $f$ is differentiable at $z_0$ if the following limit exists: $\lim _{z \to z_0} \frac {f(z) - f(z_0)}{z-z_0}$ and this limit is denoted by $f'(z_0)$ , the derivative of $f$ at $z_0$ , i.e., $f'(z_0) = \lim _{z \to z_0} \frac {f(z) - f(z_0)}{z-z_0}$ An equivalent alternative formulation of the derivative is $f'(z_0) = \lim _{h \to 0} \frac {f(z_0 +h) -f(z_0)}{h}$ In the latter formulation, it is important to note that $h\in \C$ .

example 1 Let $z_0 \in \C$ . Compute $f'(z_0)$ for $f(z) = z^4$ .
Since $z^4 - z_0^4$ factors readily, we will use the first formulation of the derivative: $f'(z_0) =\lim _{z \to z_0} \frac {z^4 - z_0^4}{z-z_0} =\lim _{z \to z_0} \frac {(z-z_0)(z+z_0)(z^2+z_0^2)}{z-z_0}$ $= \lim _{z \to z_0} (z+z_0)(z^2+z_0^2) = 2z_0\cdot 2z_0^2 = 4z_0^3$ Since this formula for $f'$ holds for any $z_0 \in \C$ , we can write, $f'(z) = 4z^3$

(problem 1a) Use the definition of the derivative to compute $f'(z)$ for the following functions: \begin{align*} i) \;\; &f(z) = c \quad f'(z) =\answer{0}\\ ii) \;\; &f(z) = z^2 \quad f'(z) =\answer{2z}\\ iii) \;\; &f(z) = z^3 \quad f'(z) =\answer{3z^2}\\ iv) \;\; & f(z) = \frac{1}{z}, (z \neq 0) \quad f'(z) =\answer{-\frac{1}{z^2}} \end{align*}

Here is a video solution to problem 1a, part iv:

Power Rule If $n$ is a positive integer, then $f(z) = z^n$ is differentiable on $\C$ and $f'(z) = \frac {d}{dz} z^n = nz^{n-1}$

Proof: We will use the factorization $z^n - z_0^n = (z-z_0)\left (z^{n-1} + z^{n-2}z_0 + \cdots + zz_0^{n-2} + z_0^{n-1} \right )$ We have \begin{align*} f'(z_0) &= \lim _{z \to z_0} \frac{z^n - z_0^n}{z-z_0} \\ &=\lim _{z \to z_0} \frac{(z-z_0)\left (z^{n-1} + z^{n-2}z_0 + \cdots + zz_0^{n-2} + z_0^{n-1} \right )}{z-z_0} \\ &= \lim _{z \to z_0} \left (z^{n-1} + z^{n-2}z_0 + \cdots + zz_0^{n-2} + z_0^{n-1} \right )\\ & = z_0^{n-1} + z_0^{n-1} + \cdots z_0^{n-1}\\ & = nz_0^{n-1} \end{align*}
Since this holds for all $z_0 \in \C$ , we can write $f'(z) = nz^{n-1}$ . $\blacksquare$

The power rule can also be proved using the alternate formulation of the derivative with the help of the Binomial Theorem: $(z_0 + h)^n = \sum _{k = 0}^n \binom {n}{k} z_0^{n-k} h^k$

(problem 1b, root of unity) Use the factorization $z^n - z_0^n = (z-z_0)\left (z^{n-1} + z^{n-2}z_0 + \cdots + zz_0^{n-2} + z_0^{n-1} \right )$ to show that if $\alpha \neq 1$ is an $n^{th}$ root of unity, i.e., $\alpha ^n = 1$ , then $\alpha ^{n-1} + \alpha ^{n-2} + \cdots + \alpha + 1 = 0$

(problem 1c, power rule) Prove that for $z_0 \neq 0$ , the power rule holds for negative integers.

Here is a video solution to problem 1c:

If a function is differentiable at $z= z_0$ , then it must be continuous there.

If $f$ is differentiable at $z = z_0$ then $f$ is continuous at $z = z_0$

Proof: $\lim _{z\to z_0} \left [f(z) - f(z_0)\right ] = \lim _{z\to z_0} \frac {f(z) - f(z_0)}{z-z_0} \cdot (z-z_0)$ $= \lim _{z\to z_0} \frac {f(z) - f(z_0)}{z-z_0} \cdot \lim _{z\to z_0}(z-z_0) = f'(z_0) \cdot 0 = 0$ Thus $\lim _{z\to z_0} f(z) = f(z_0),$ and so $f(z)$ is continuous at $z = z_0$ . $\blacksquare$

We next state familiar rules for differentiation that extend from functions of a real variable to complex functions.

Basic Rules for Differentiation If $f$ and $g$ are complex functions that are differentiable at $z = z_0$ then the following hold: \begin{align*} \mbox{Sum Rule} & \quad (f+g)'(z_0) = f'(z_0) + g'(z_0)\\[6 pt] \mbox{Product Rule} & \quad (fg)'(z_0) = f'(z_0)g(z_0) + f(z_0)g'(z_0)\\[6pt] \mbox{Quotient Rule} &\quad \left (\frac{f}{g}\right )'(z_0) = \frac{f'(z_0)g(z_0) - f(z_0)g'(z_0)}{g^2(z_0)}\quad (\mbox{if} \; g(z_0) \neq 0) \end{align*}

Proof: We prove the quotient rule and leave the others as exercises. If $g(z_0) \neq 0$ then $\left (\frac {1}{g}\right )'(z_0) = \lim _{z \to z_0} \frac {\frac {1}{g(z)} - \frac {1}{g(z_0)}}{z-z_0}$ $= \lim _{z \to z_0} \frac {g(z_0)- g(z)}{g(z)\cdot g(z_0) \cdot (z-z_0)}$ $= \lim _{z \to z_0} \frac {g(z_0)- g(z)}{z-z_0} \cdot \lim _{z \to z_0} \frac {1}{g(z)\cdot g(z_0)}$ $= -g'(z_0) \cdot \frac {1}{g^2(z_0)} = -\frac {g'(z_0)}{g^2(z_0)}$ The quotient rule follows from combining this reciprocal rule with the product rule. $\blacksquare$

Chain Rule If $g(z)$ is differentiable at $z = z_0$ and $f(z)$ is differentiable at $z = g(z_0)$ , then $f \circ g$ is differentiable at $z_0$ and $(f\circ g)'(z_0) = \frac {d}{dz} f(g(z)) \Big |_{z=z_0} = f'(g(z_0)) \cdot g'(z_0)$

Proof: Let $\phi (z) = \left \{ \begin {array}{lr} \frac {f(g(z)) - f(g(z_0))}{g(z) -g(z_0)} & \ \text {for} \ g(z) \neq g(z_0) \\[12pt] f'(g(z_0)) & \ \text {for} \ g(z) = g(z_0) \end {array} \right .$ Then $\lim _{z \to z_0} \phi (z) = f'(g(z_0))$ Now, \begin{align*} (f\circ g)'(z_0) &= \lim _{z\to z_0} \frac{f(g(z)) - f(g(z_0))}{z-z_0} \\[10pt] &= \lim _{z \to z_0} \phi (z_0) \cdot \frac{g(z) - g(z_0)}{z-z_0} \\[10pt] &= f'(g(z_0)) \cdot g'(z_0) \end{align*} $\blacksquare$

example 2 Find the derivative of $f(z) = (z^2 + iz + 1)^5$ .
By the Chain Rule and the Power rule, we have $f'(z) = 5(z^2 + z + 1)^4 \cdot (2z +i)$

(problem 2) Find the derivative of $f(z) = (z^3 - 2iz^2 + 3 + 2i)^4$ .
$f'(z) = \answer {4(z^3 - 2iz^2 + 3 + 2i)^3 (3z^2 -4iz)}$

In the next example, we will see a function that is differentiable at $0$ but nowhere else.

example 3 Let $f(z) = \overline {z}^2$ . Show that $f'(0) = 0$ but that $f$ is not differentiable for any $z_0 \neq 0$ .
We begin with the derivative at $0$ : $f'(0) = \lim _{z \to 0} \frac {\overline {z}^2}{z} = \lim _{z \to 0} \frac {\overline {z}^3}{|z|^2}$ We will separate this into real and imaginary parts using \begin{align*} \overline{z}^3 &= (x -iy)^3 = x^3 - 3x^2(iy) + 3x(iy)^2 - (iy)^3 &= (x^3 -3xy^2) + i(y^3 - 3x^2y) \end{align*}

Doing so gives \begin{align*} f'(0) &= \lim _{z \to 0} \frac{\overline{z}^3}{|z|^2}\\ &= \lim _{(x,y) \to (0,0)} \frac{x^3 -3xy^2}{x^2 +y^2} + i\cdot \lim _{(x,y) \to (0,0)} \frac{y^3 - 3x^2y}{x^2 +y^2}\\ & = 0 + 0i \;\; \mbox{(verify)} \end{align*}

Next, suppose $z_0 \neq 0$ , then the alternative formulation of $f'(z_0)$ gives \begin{align*} f'(z_0) &= \lim _{h \to 0} \frac{\left (\overline{z_0 + h}\right )^2-\overline{z_0}^2 }{h}\\[6pt] &=\lim _{h \to 0} \frac{\overline{z_0}^2 +2\overline{z_0h}+ \overline{h}^2 -\overline{z_0}^2 }{h}\\[6pt] &=\lim _{h \to 0} \frac{\overline{h} \left (2\overline{z_0}+ \overline{h}\right )}{h}. \end{align*}

This last limit is equal to $2\overline {z_0}$ if $h$ is real and it is equal to $-2z_0$ if $h$ is purely imaginary. Since $2\overline {z_0} \neq -2\overline {z_0}$ if $z_0 \neq 0$ , this limit does not exist and $f$ is not differentiable at $z_0 \neq 0$ .
Hence $f(z) = \overline {z}^2$ is only differentiable at $0$ and $f'(0) = 0$

(problem 3) Show that the function $f(z) = \overline {z}$ is nowhere differentiable.

Here is a video solution to problem 3:

2024-09-27 14:06:57