Complex Functions

In general, the definition of a Derivative holds for complex functions defined on $[a, b]$ .

Theorem (Differentiable is Continuous)
Theorem (Derivative Rules) also hold (with the same proofs). For complex function $f$ , the real $f_{1}$ and imaginary $f_{2}$ parts can be described as:

f (t) = f_{1} (t) + i f_{2} (t)

for $t \in [a, b]$ and as $f_{1}, f_{2}$ are real,

f^{'} (x) = f_{1}^{'} (x) + i f_{2}^{'} (x)

$f$ is differentiable at $x$ iff $f_{1}, f_{2}$ are both differentiable at $x$ .

Lagrange Mean Value Theorem Fails

The Theorem (Lagrange Mean Value Theorem) does not hold. For some real $x$ ,

f (x) = e^{i x} = cos x + i sin x

Then $f (2 π) = f (0) = 1 - 1 = 0$ , but $f^{'} (x) = i e^{i x}$ and $∣ f^{'} (x) ∣ = 1$ for all real $x$ . So,

1 = f^{'} (x) = \frac{f ( 2 π ) - f ( 0 )}{2 π - 0} = 0

which is a contradiction.

On the segment $(0, 1)$ , let $f (x) = x$ and

g (x) = x + x^{2} e^{i / x^{2}}

Since $∣ e^{i t} ∣ = 1$ for all real $t$ , we see that

x \to 0 lim \frac{f ( x )}{g ( x )} = 1

But then

g^{'} (x) = 1 + (2 x - \frac{2 i}{x}) e^{i / x^{2}}

for $x \in (0, 1)$ and that

∣ g^{'} (x) ∣ \geq 2 x - \frac{2 i}{x} - 1 \geq \frac{2}{x} - 1 = \frac{2 - x}{x}

Hence

\frac{f ^{'} ( x )}{g ^{'} ( x )} = \frac{1}{∣ g ^{'} ( x ) ∣} \leq \frac{x}{2 - x}

and so

x \to 0 lim \frac{f ^{'} ( x )}{g ^{'} ( x )} = 0