MAT334-2018F > MAT334--Lectures & Home Assignments

3.2 Q6

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**Kris**:

Can someone help me solve this problem?

**yunhao guan**:

Here is how I think about this question.

Let $h = f-g$, and we know that h is analytic. Therefore $\left|e^ {f-g} \right| = e^ {Reh} = e^0 = 1$, since $Ref = Reg$

So $\left|e^ {f-g} \right| = 1 $ and it implies that $e^ {f-g}$ is constant which means $e^ {f-g} = c$ so $f-g = ln(c)$

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