# 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, 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 ProofOfComplexMeanvalueTheorem 2013-03-22 14:34:39 2013-03-22 14:34:39 Wolfgang (5320) Wolfgang (5320) 21 Wolfgang (5320) Proof msc 26A06