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Complex mean value theorem: Let $g$ be a holomorphic function defined on an open convex subset $D_{g}$ of $ℂ$. Let $v$ and $u$ be two distinct points in $D_{g}$. Then there exist $z₁,z₂∈(u,v)$ such that

$Re(g′(z₁))=Re(((g(u)-g(v))/(u-v)))$,

$Im(g′(z₂))=Im(((g(u)-g(v))/(u-v)))$, i.e.,

$g(u)=g(v)+(u-v)(Re(g′(z₁))+iIm(g′(z₂)))$.

In fact, we have $z₁=v+t₁(u-v),t₁∈(0,1)$ and $z₂=v+t₂(u-v),t₂∈(0,1)$ from the proof of this Theorem.

My question is about the uniqueness of $t₁,t₂∈(0,1)$ . The proof of the result indicate only their existence.

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1 Answer 1

up vote 3 down vote accepted

Let's consider $g(z) := z$ for $z \in \mathbb{C}$. We are looking for $z_1,z_2 \in (u,v)$ such that

$$\begin{align*} \text{Re} \, g'(z_1)=1 \stackrel{!}{=} \text{Re} \left(\frac{g(u)-g(v)}{u-v}\right) = 1 \tag{1}\\ \text{Im} \, g'(z_2)=0 \stackrel{!}{=} \text{Im} \left(\frac{g(u)-g(v)}{u-v}\right) = 0 \tag{2}\end{align*} $$

This shows that all $z_1, z_2 \in (u,v)$ fulfill the given equations.

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Thank you. But I am asking on the unicity of $t₁,t₂∈(0,1)$. No matter with $z₁,z₂∈(v,u)$. –  ZE1 Dec 16 '12 at 12:57
    
But you defined $t_1$ by $z_1=v + t_1 \cdot (u-v)$. Since the equality (1) holds for all $z_1 \in (u,v)$ you can choose $t_1 \in (0,1)$ arbritary. Same for $t_2$. –  saz Dec 16 '12 at 13:03
    
Yes. thank you very much. –  ZE1 Dec 16 '12 at 13:07

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