proof of complex mean-value theorem

The function $h(t)=\operatorname{Re}\frac{f(a+t(b-a))-f(a)}{b-a}$ is a function defined on [0,1]. We have $h(0)=0$ and $h(1)=\operatorname{Re}\frac{f(b)-f(a)}{b-a}$ . By the ordinary mean-value theorem, there is a number $t$ , $0<t<1$ , such that $h^{\prime}(t)=h(1)-h(0)$ . To evaluate $h^{\prime}(t)$ , we use the assumption that $f$ is complex differentiable (holomorphic). The derivative of $\frac{f(a+t(b-a))-f(a)}{b-a}$ is equal to $f^{\prime}(a+t(b-a))$ , then $h^{\prime}(t)=\operatorname{Re}(f^{\prime}(a+t(b-a)))$ , so $u=a+t(b-a)$ satisfies the required equation. The proof of the second assertion can be deduced from the result just proved by applying it to the function f multiplied by i.

Title	proof of complex mean-value theorem
Canonical name	ProofOfComplexMeanvalueTheorem
Date of creation	2013-03-22 14:34:39
Last modified on	2013-03-22 14:34:39
Owner	Wolfgang (5320)
Last modified by	Wolfgang (5320)
Numerical id	21
Author	Wolfgang (5320)
Entry type	Proof
Classification	msc 26A06