Find all the solutions of $e^z=i$
I don't know where to start.
I want to do $z=\ln(i)$, but have no idea where that would lead me.
Thanks in advance.
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Hint: write $z=x+iy$ and solve the real and imaginary parts separately: $$e^{x+iy} = e^x (\cos y + i \sin y) = 0 + i$$
Adding more detail:
$$\begin{cases}e^x \cos y = 0 \\ e^x \sin y = 1\end{cases}$$ Noting that $e^x > 0$ for any $x$, $$e^x \cos y = 0 \implies \cos y = 0 \implies y = \pi/2 + k \implies \sin y = \pm 1$$ Again, using the fact that $e^x>0$, this implies $$\sin y = 1 \implies \begin{cases}x=0 \\ y=\pi/2 + 2\pi k\end{cases}$$
angryavian
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Okay, I've got that. But now what do I need to do? Take natural logs to both sides? $\cos y+i\sin y=\ln(i)$. – Mr Croutini Jan 07 '14 at 15:22
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As $\displaystyle i=\cos\frac\pi2+i\sin\frac\pi2$
using Euler formula $\displaystyle i=e^{i\frac\pi2}=e^{i\left(2m\pi+\frac\pi2\right)}$ where $m$ is any integer
lab bhattacharjee
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@MrCroutini, what are the values of $\cos\frac\pi2,\sin\frac\pi2$? To derive the relation we can set $0=r\sin\psi, 1=r\cos\psi$ where $r>0$ – lab bhattacharjee Jan 07 '14 at 15:26
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$\cos \frac{\pi}{2}=0$ and $\sin \frac{\pi}{2}=1$. Okay, I understand that, thanks. – Mr Croutini Jan 07 '14 at 15:28
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I understand everything, apart from the last equality. Could you explain it? Is it because $\sin(x)$ is $2\pi$ periodic? – Mr Croutini Jan 07 '14 at 15:42
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@MrCroutini, right. http://math.stackexchange.com/questions/3861/what-is-your-favorite-proof-that-eix-has-a-period-of-2-pi – lab bhattacharjee Jan 07 '14 at 15:44
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Write $i$ as ${\large e^{i\left(\dfrac\pi2 + 2n\pi\right)}}$. Then compare the powers on both sides, take LCM and boom!
The Demonix _ Hermit
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