# Toronto Math Forum

## MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Kris on December 03, 2018, 11:52:10 PM

Title: 3.2 Q6
Post by: Kris on December 03, 2018, 11:52:10 PM
Can someone help me solve this problem?
Title: Re: 3.2 Q6
Post by: yunhao guan 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 $f-g = ln(c)$