# Express $1/5i$ into polar form

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

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?

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

$$\dfrac{-\mathrm{i}}{5} = -\mathrm{i} \cdot \frac{1}{5}$$. 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?
$$z=0+i\left(-\frac{1}{5}\right)$$ 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?