### Author Topic: 3.2 Q6  (Read 1033 times)

#### Kris

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##### 3.2 Q6
« on: December 03, 2018, 11:52:10 PM »
Can someone help me solve this problem?

#### yunhao guan

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##### Re: 3.2 Q6
« Reply #1 on: December 04, 2018, 12:22:07 AM »
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 $f-g = ln(c)$