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Question Number 39914 by solihin last updated on 13/Jul/18

Commented by MrW3 last updated on 13/Jul/18

$$z=\frac{i}{2-i}=\frac{i(2+i)}{(2-i)(2+i)}=\frac{-1+2i}{5}=-\frac{1}{5}+\frac{2}{5}i$$$$=\frac{1}{\sqrt{5}}(-\frac{1}{\sqrt{5}}+\frac{2}{\sqrt{5}}i)$$$$=r(\cos \theta +i\sin \theta )$$$$withr=\frac{1}{\sqrt{5}}$$$$\theta =\pi -{\tan }^{-1}2\approx 116.6°$$

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