2.5: Derivatives

The definition of the complex derivative of a complex function is similar to that of a real derivative of a real function: For a function $f(z)$ the derivative $f$ at $z_0$ is defined as

$$f'(z_0) = \lim_{z \to z_0} \dfrac{f(z) - f(z_0)}{z - z_0} \nonumber$$

Provided, of course, that the limit exists. If the limit exists we say $f$ is analytic at $z_0$ or $f$ is differentiable at $z_0$.

Remember: The limit has to exist and be the same no matter how you approach $z_0$!

If $f$ is analytic at all the points in an open region $A$ then we say $f$ is analytic on $A$.

As usual with derivatives there are several alternative notations. For example, if $w = f(z)$ we can write

$$f'(z_0) = \dfrac{dw}{dz} \lvert_{z_0} = \lim_{z \to z_0} \dfrac{f(z) - f(z_0)}{z - z_0} = \lim_{\Delta \to 0} \dfrac{\Delta w}{\Delta z} \nonumber$$

Example $\PageIndex{1}$

Find the derivative of $f(z) = z^2$.

Solution

We did this above in Example 2.2.1. Take a look at that now. Of course, $f'(z) = 2z$.

Example $\PageIndex{2}$

Show $f(z) = \overline{z}$ is not differentiable at any point $z$.

Solution

We did this above in Example 2.2.2. Take a look at that now.

Challenge. Use polar coordinates to show the limit in the previous example can be any value with modulus 1 depending on the angle at which $z$ approaches $z_0$.

Derivative rules

It wouldn’t be much fun to compute every derivative using limits. Fortunately, we have the same differentiation formulas as for real-valued functions. That is, assuming $f$ and $g$ are differentiable we have:

• Sum rule: $$\dfrac{d}{dz} (f(z) + g(z)) = f' + g' \nonumber$$
• Product rule: $$\dfrac{d}{dz} (f(z) g(z)) = f'g + fg' \nonumber$$
• Quotient rule: $$\dfrac{d}{dz} (f(z)/g(z)) = \dfrac{f'g - fg'}{g^2} \nonumber$$
• Chain rule: $$\dfrac{d}{dz} g(f(z)) = g'(f(z)) f'(z) \nonumber$$
• Inverse rule: $$\dfrac{df^{-1} (z)}{dz} = \dfrac{1}{f' (f^{-1} (z))} \nonumber$$

To give you the flavor of these arguments we’ll prove the product rule.

$$\begin{array} {rcl} {\dfrac{d}{dz} (f(z) g(z))} & = & {\lim_{z \to z_0} \dfrac{f(z) g(z) - f(z_0) g(z_0)}{z - z_0}} \\ {} & = & {\lim_{z \to z_0} \dfrac{(f(z) - f(z_0)) g(z) + f(z_0) (g(z) - g(z_0))}{z - z_0}} \\ {} & = & {\lim_{z \to z_0} \dfrac{f(z) - f(z_0)}{z - z_0} g(z) + f(z_0) \dfrac{(g(z) - g(z_0))}{z - z_0}} \\ {} & = & {f'(z_0) g(z_0) + f(z_0) g'(z_0)} \end{array} \nonumber$$

Here is an important fact that you would have guessed. We will prove it in the next section.

Theorem $\PageIndex{1}$

If $f(z)$ is defined and differentiable on an open disk and $f'(z) = 0$ on the disk then $f(z)$ is constant.