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Find the argument of $\displaystyle \pi e^{-\frac{3i\pi}{2}}$

I thought the formula for the argument was $\arg{z} = i\log\frac{|z|}{z}$

In this case $|z|= \pi$, so it turns out that $-i \log(e^{-3i\pi/2}) = \frac{\pi}{2}.$

However, I don't understand why $-i \log(e^{-3i\pi/2}) = \frac{\pi}{2}$.

Could someone please explain why.

Also, is there a quicker method?

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The argument is the angle that the vector makes with respect to the positive real axis.

Using that $e^{i \pi} = \cos \pi + i \sin \pi$ and that $-\frac{3 \pi } {2} = \frac{\pi}{2}$, we see then that

$$ \textrm{exp} \left(\frac{3\pi i } {2}\right) = \textrm{exp} \left(\frac{\pi i}{2}\right) = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2}$$

Thus the argument (angle) is $\frac{\pi}{2}$.

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  • $\begingroup$ Thanks. So you added $2\pi$ to $-\frac{3}{2}\pi$ to put it in the range where the argument is defined, is that correct? $\endgroup$ – GGG Jan 9 '16 at 5:42
  • $\begingroup$ The argument is not unique: it is only defined modulo $2 \pi$. Both are correct. If you want a unique one, you are looking for the principal argument. $\endgroup$ – Future Jan 9 '16 at 5:45
  • $\begingroup$ You missed $i$ in the second exponential. $\endgroup$ – Claude Leibovici Jan 9 '16 at 6:07
  • $\begingroup$ I fixed it thanks! $\endgroup$ – Future Jan 9 '16 at 15:18
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See $π$ is the radius so it doesnt matter in so lets start our complex number is $$\frac{π}{e^{i3π/2}}=\frac{π}{-i}=0+(-π.)(i)$$ after rationalising . So now the theta becomes as it is third quadrant $\theta=π+tan^-{1}((-π)/0)=π-π/2=π/2$ thats all and hint $log(x+iy)=log(i\theta)+|z|$ where $\theta$ is its argument and z is its radius from origin.

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