Question Number 39914 by solihin last updated on 13/Jul/18
Commented by MrW3 last updated on 13/Jul/18
$${z}=\frac{{i}}{\mathrm{2}−{i}}=\frac{{i}\:\left(\mathrm{2}+{i}\right)}{\left(\mathrm{2}−{i}\right)\left(\mathrm{2}+{i}\right)}=\frac{−\mathrm{1}+\mathrm{2}{i}}{\mathrm{5}}=−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{5}}{i} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}{i}\right) \\ $$$$={r}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right) \\ $$$${with}\:{r}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\theta=\pi−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\:\approx\mathrm{116}.\mathrm{6}° \\ $$