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Question Number 95121 by i jagooll last updated on 23/May/20

Commented by bobhans last updated on 23/May/20

$$\mathit{z}=\frac{5}{2+\mathit{i}}\times \frac{2-\mathit{i}}{2-\mathit{i}}=\frac{10-5\mathit{i}}{4+1}=2-\mathit{i}$$$$\mathit{a}\mathit{r}\mathit{g}\mathit{u}\mathit{m}\mathit{e}\mathit{n}\mathit{t}\mathit{o}\mathit{f}\mathit{z}\Rightarrow \phi ={\tan }^{-1}\left(\frac{-1}{2}\right)$$

Answered by mathmax by abdo last updated on 24/May/20

$$\mathrm{z}=\frac{5}{2+\mathrm{i}}\mathrm{we}\mathrm{have}2+\mathrm{i}=\sqrt{5}(\frac{2}{\sqrt{5}}+\frac{\mathrm{i}}{\sqrt{5}})=\sqrt{5}(\cos \theta +\mathrm{isin}\theta )\Rightarrow$$$$\mathrm{r}=\sqrt{5}\mathrm{and}\theta =\arctan \left(\frac{1}{2}\right)\Rightarrow \mathrm{z}=\frac{5}{\sqrt{5}{\mathrm{e}}^{\mathrm{iarctan}\left(\frac{1}{2}\right)}}=\sqrt{5}{\mathrm{e}}^{-\mathrm{iarctan}\left(\frac{1}{2}\right)}\Rightarrow$$$$\arg \left(\mathrm{z}\right)\equiv \arctan (-\frac{1}{2})\left[2\pi \right]$$

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