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$$ \cfrac{1}{5i} = - \cfrac{i}{5} = \cfrac{1}{5} e^{\frac{3\pi i}{2}} $$

I know to convert a standard equation into polar form but this one is confusing me somewhat.

First step is multiply by $i$ so it becomes $-i/5$

After that I’m stuck. I don't know how to find the angle from this. I would assume the real is $0$ but that doesn't match the answer provided which is $\cfrac{3 \pi}{2}$ or $\cfrac{- \pi}{2}$

Any ideas?

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  • $\begingroup$ Why doesn't an argument of $3\pi/2$ (or for that matter $-\pi/2$) match having real part $0$?? $\endgroup$ – Henning Makholm Jun 14 at 0:36
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    $\begingroup$ Do you know how to express $i$ or $-i$ in polar form? $\endgroup$ – J. W. Tanner Jun 14 at 0:39
  • $\begingroup$ Care to explain please? $\endgroup$ – Mr A Jun 14 at 0:39
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    $\begingroup$ What is question $ \cfrac{i}{5}$ or $ \cfrac{1}{5 i}$ ? $\endgroup$ – Ajay Mishra Jun 14 at 0:43
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    $\begingroup$ Draw a picture. Ask yourself what is the angle with the positive real axis. $\endgroup$ – GEdgar Jun 14 at 0:44
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$\dfrac{-\mathrm{i}}{5} = -\mathrm{i} \cdot \frac{1}{5}$. What's the angle and magnitude for $-\mathrm{i}$? What's the angle and magnitude for $\frac{1}{5}$? What's the product of those two magnitudes and the sum of those two angles?

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$$z=0+i\left(-\frac{1}{5}\right)$$ Now the problem is, what is $\text{Arg} (z)$. It is the unique value that satisfies $$-\pi < \text{arg}(z) \leq \pi$$ where $\text{arg}(z)$ is the usual argument of given $z$. Here $\text{Arg}(z)=- \frac{\pi}{2}$. Can you see why?

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