# Toronto Math Forum

## MAT334--2020F => MAT334--Tests and Quizzes => Test 1 => Topic started by: Milan Miladinovic on October 14, 2020, 03:42:15 PM

Title: 2020S-TT1 Q1
Post by: Milan Miladinovic on October 14, 2020, 03:42:15 PM
I'm having trouble understanding where the $-1+i$ term comes from in the following line:
$\dfrac{e^{3z} - e^{-3z}}{e^{3z} + e^{-3z}} = 1 + 2i \implies e^{6z} = -1 + i$.

I've tried the following:
\begin{align*} \dfrac{e^{3z} - e^{-3z}}{e^{3z} + e^{-3z}} &= 1 + 2i\\ \dfrac{e^{6z} - 1}{e^{6z} + 1} &= 1 + 2i\\ e^{6z} - 1 &= (1 + 2i)(e^{6z} + 1)\\ e^{6z} &= (1 + 2i)(e^{6z} + 1) + 1 \end{align*}

How do we get from $(1 + 2i)(e^{6z} + 1) + 1$ to $-1 + i$? Have I done something wrong somewhere in my calculation?
Title: Re: 2020S-TT1 Q1
Post by: Maria-Clara Eberlein on October 14, 2020, 03:52:26 PM
This is how I got the answer, hope this helps!
Title: Re: 2020S-TT1 Q1
Post by: Milan Miladinovic on October 14, 2020, 04:11:40 PM
Awesome, that makes sense! I was overthinking it