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LEt $f:\mathbb{D}\rightarrow\mathbb{C}$ be an holomorphic while $\mathbb{D}$ is convex set and let $Ref'(z) >0$ show that for every $z\in D$ $\quad$$f(z)$ is one to one function.

I was thinking on using the cauchy reiman eqations \begin{bmatrix}\dfrac{\partial u}{\partial x}&\dfrac{\partial u}{\partial y}\\-\dfrac{\partial u}{\partial y}&\dfrac{\partial u}{\partial x}\end{bmatrix}

Now this matrix is reversible since $Ref'(z) >0$ and the determinant is not zero. now I wanted to show in some way that $f(z_1) = f(z_2) \rightarrow z_1 = z_2$ I am trying to think how should I use the fact that $D$ is convex set can help me here.

Also, why this is not the case anymore when $D$ is startlike domain? namely , there exist $z_0 \in D$ such that for every $z \in D$ and $t \in (0,1)$ $$tz_1 +(1-t)z_2 \in D$$

Sagigever
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1 Answers1

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Assume that $a,b\in\mathbb D$, $a\ne b$ and $f(a)=f(b)$. Then $$ 0=f(b)-f(a)=\int_0^1 \frac{d}{dt}f\big(a+t(b-a)\big)\,dt =\int_0^1 f'\big(a+t(b-a)\big)\cdot(b-a)\,dt \\ =(b-a)\int_0^1 f'\big(a+t(b-a)\big)\,dt $$ and thus $$ \int_0^1 f'\big(a+t(b-a)\big)\,dt=0 \quad\Longrightarrow\quad \int_0^1 {\mathrm{Re}}\,f'\big(a+t(b-a)\big)\,dt=0. $$ Contradiction.

Yiorgos S. Smyrlis
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