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I want to prove the following theorem

If $f$ is continuous on $\mathbb{C}$ and analytic in $\{ z\mid \operatorname{Im}(z) \neq 0\}$, then $f$ is analytic in $\mathbb{C}$.

First I know from Morera's theorem, continuous function $f$ becomes analytic if $\int_\Gamma f(z)dz=0$ for any closed contour $\Gamma$ in the domain $D$.

Since $\operatorname{im}(z)=0$ implies $z \in \mathbb{R}$, I see $f$ is analytic in $\mathbb{C} \setminus \mathbb{R}$ but want to extend this to $\mathbb{C}$.

I saw A continuous function $f$ is analytic everywhere except along a simple closed contour $C$ in domain $D$, then $f$ is analytic everywhere in $D$. this post, but for me, it seems too complicated. I think there is a simple way to prove or show this theorem. Maybe my approach with using Morera's theorem is a bad starting point.

Thomas Andrews
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phy_math
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1 Answers1

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A proof using Morera's theorem should go through without problems.

To prove that the value of something is zero, it is enough to show that it's absolute value is lesser than all $\epsilon>0$.

So, divide a rectangular contour into 3 rectangular parts. A part above the real axis, say $Im(z)\geq \delta$ , the part below, say $Im(z)\leq -\delta$ and a tiny rectangle that has a side above the real axis and one below, $Im(z)\in [-\delta,\delta]$.

The function $f$ is bounded on the whole rectangular contour because of compactness. Therefore, the integral on the tiny rectangle must be negligible. This forces the whole integral on all three rectangles to be lesser than any $\epsilon>0$, using modulus of integral bounds and triangle inequality.

Umesh Shankar
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