Jovany Clayton

2022-07-13

What is ${\mathrm{tan}}^{-1}\left(5i/3\right)$
My progress: Let $\mathrm{tan}x=\frac{5i}{3}=\frac{\mathrm{sin}x}{\mathrm{cos}x}$ I tried using $\mathrm{sin}x=\frac{{e}^{ix}-{e}^{-ix}}{2},\mathrm{cos}x=\frac{{e}^{ix}+{e}^{-ix}}{2}$ to show that $\frac{{e}^{ix}-{e}^{-ix}}{{e}^{ix}+{e}^{-ix}}=\frac{-5}{3}$ or ${e}^{2ix}=\frac{-1}{4}$, but I'm stuck here.

Kroatujon3

Expert

You are fine. Just solve ${e}^{2ix}=\frac{-1}{4}$, taking into account that x is complex. Write x=r+ic to get ${e}^{-2c+2ir}=\frac{-1}{4}=\frac{1}{4}{e}^{i\pi }$ and identify $r=\pi /2$ and $c=\mathrm{ln}2$. So
$x=\pi /2+i\mathrm{ln}2$

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