proof of complex mean-value theorem

The function $h(t)=\mathrm{Re}\frac{f(a+t(b-a))-f(a)}{b-a}$ is a function defined on [0,1].We have $h(0)=0$ and $h(1)=\mathrm{Re}\frac{f(b)-f(a)}{b-a}$.By the ordinary mean-value theorem, there is a number $t$, , 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 $h^{\prime }(t)=\mathrm{Re}(f^{\prime }(a+t(b-a)))$, 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.