2.5: Derivatives

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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.