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Let $f$ be an entire function and $B$ be a bounded open set in $\mathbb {C} $. Prove that boundary of image of $B$ under $f$ is contained in image of boundary of $B$. Does the same result is true for unbounded open set in $\mathbb C.$

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If $f$ is constant it is clear. Assume it is non-constant.

Then by the open mapping theorem it is open. Therefore, interior points of $B$ are mapped to interior points of $f(B)$. Hence, boundary points of $f(B)$ can only be image of boundary points of $B$. In fact, if $w\in\partial f(B)$ there is a sequence $w_n=f(z_n)$ with $z_n\in B$ such that $w_n\to w$. Since $B$ is bounded $z_n$ have an accumulation point $z$, which must be in the boundary of $B$.

For the unbounded case we can consider $B=\{z\in\mathbb{C}:\ -2\pi<Im(z)<2\pi\}$ and $f(z)=e^z$.

Then $f(B)$ is equal to the $\mathbb{C}\setminus\{0\}$.

The origin is a boundary point of $f(B)$. But $e^z$ doesn't map any point of the plane to the origin, and in particular no point of the boundary of $B$.

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  • $\begingroup$ ok thanking you. Nice explanation. $\endgroup$ – neelkanth Apr 28 '15 at 14:40

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