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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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Test 1 / 2020S-TT1 Q1
« 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?
$\dfrac{e^{3z} - e^{-3z}}{e^{3z} + e^{-3z}} = 1 + 2i \implies e^{6z} = -1 + i$.
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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Test 1 / Re: 2020 Night Sitting #1
« on: October 14, 2020, 02:52:16 PM »
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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