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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=(i/(2−i))=((i (2+i))/((2−i)(2+i)))=((−1+2i)/5)=−(1/5)+(2/5)i  =(1/(√5))(−(1/(√5))+(2/(√5))i)  =r(cos θ+i sin θ)  with r=(1/(√5))  θ=π−tan^(−1) 2 ≈116.6°

$${z}=\frac{{i}}{\mathrm{2}−{i}}=\frac{{i}\:\left(\mathrm{2}+{i}\right)}{\left(\mathrm{2}−{i}\right)\left(\mathrm{2}+{i}\right)}=\frac{−\mathrm{1}+\mathrm{2}{i}}{\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}}{i} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\left(−\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}+\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}{i}\right) \\ $$$$={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${with}\:{r}=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}} \\ $$$$\theta=\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\:\approx\mathrm{116}.\mathrm{6}° \\ $$

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