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**Test 1 / Re: 2020S-TT1 Q1**

« **on:**October 14, 2020, 04:11:40 PM »

Awesome, that makes sense! I was overthinking it

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Awesome, that makes sense! I was overthinking it

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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?

$\dfrac{e^{3z} - e^{-3z}}{e^{3z} + e^{-3z}} = 1 + 2i \implies e^{6z} = -1 + i$.

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?

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I got 2 complex roots as well. It looks like there's a slight typo, should be $e^z = \dfrac{-1 \pm i\sqrt{3}}{2}$, so we get the same answer as the solution: $log\left(\dfrac{-1 \pm i\sqrt{3}}{2}\right) = \left(\pm\dfrac{2}{3} + 2n\right)i\pi,$ for $n\in\mathbb{Z}$.

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