7.1: Complex Continuity and Differentiability
The concept of a continuous complex function makes use of an “epsilon-delta definition”, similar to the definition for functions of real variables (see Chapter 1):
Definition: Word
A complex function \(f(z)\) is continuous at \(z_0 \in \mathbb{C}\) if, for any \(\epsilon > 0\) , we can find a \(\delta > 0\) such that \[\big|\, z - z_0 \,\big| < \delta \;\;\; \Rightarrow \;\;\; \big|\, f(z) - f(z_0) \,\big| < \epsilon.\]
Here, \(\big|\cdots\big|\) denotes the magnitude of a complex number. If you have difficulty processing this definition, don’t worry; it basically says that as \(z\) is varied smoothly, there are no abrupt jumps in the value of \(f(z)\) .
If a function is continuous at a point \(z\) , we can define its complex derivative as \[f'(z) = \frac{df}{dz} = \lim_{\delta z\rightarrow 0} \frac{f(z+\delta z) - f(z)}{\delta z}.\] This is very similar to the definition of the derivative for a function of a real variable (see Chapter 1). However, there’s a complication which doesn’t appear in the real case: the infinitesimal \(\delta z\) is a complex number, not just a real number, yet the above definition does not specify the argument of \(\delta z\) . The choice of the argument of \(\delta z\) is equivalent to the direction in the complex plane in which \(\delta z\) points, as shown in the following figure:
In principle, we might get different results from the above formula when we plug in different infinitesimals \(\delta z\) , even in the limit where \(\delta z \rightarrow 0\) and even though \(f(z)\) is continuous.
Example \(\PageIndex{1}\)
Consider the function \(f(z) = z^*\) . According to the formula for the complex derivative, \[\lim_{\delta z \rightarrow0} \frac{f(z+\delta z) - f(z)}{\delta z} = \lim_{\delta z \rightarrow0} \frac{z^*+\delta z^* - z^*}{\delta z} = \lim_{\delta z \rightarrow0} \frac{\delta z^*}{\delta z}.\] But if we plug in a real \(\delta z\) , we get a different result than if we plug in an imaginary \(\delta z\) : \[\begin{align} \delta z \in \mathbb{R} \;\; &\Rightarrow \frac{\delta z^*}{\delta z} = 1.\\ \delta z \in i \cdot \mathbb{R} &\Rightarrow \frac{\delta z^*}{\delta z} = -1.\end{align}\]
We can deal with this complication by regarding the complex derivative as well-defined only if the above definition gives the same answer regardless of the argument of \(\delta z\) . If a function satisfies this property at a point \(z\) , we say that the function is complex-differentiable at \(z\) .
The preceding example showed that \(f(z) = z^*\) is not complex-differentiable for any \(z \in \mathbb{C}\) . On the other hand, the following example shows that the function \(f(z) = z\) is complex-differentiable for all \(z \in \mathbb{C}\) :
Example \(\PageIndex{2}\)
The function \(f(z) = z\) is complex differentiable for any \(z \in \mathbb{C}\) , since \[\lim_{\delta z \rightarrow0} \frac{f(z+\delta z) - f(z)}{\delta z} = \lim_{\delta z \rightarrow0} \frac{z+\delta z - z}{\delta z} = \lim_{\delta z \rightarrow0} \frac{\delta z}{\delta z} = 1.\] The reason the result doesn’t depend on the argument of \(\delta z\) is that the derivative formula simplifies to the fraction \(\delta z / \delta z\) , which is equal to 1 for any \(|\delta z| > 0\) . Note that we simplify the fraction to 1 before taking the limit \(\delta z \rightarrow 0\) . We can’t take the limit first, because \(0/0\) is undefined.