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Question Number 29320 by Glorious Man last updated on 07/Feb/18

$$\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{argument}\mathrm{of}\frac{-1+2\mathrm{i}}{1-3\mathrm{i}}.$$

Commented by abdo imad last updated on 08/Feb/18

$$\mid -1+2i\mid =\sqrt{5}and-1+2i=\sqrt{5}(\frac{-1}{\sqrt{5}}+i\frac{2}{\sqrt{5}})=r{e}^{i\theta }\Rightarrow$$$$r=\sqrt{5}andcos\theta =\frac{-1}{\sqrt{5}}andsin\theta =\frac{2}{\sqrt{5}}\Rightarrow tan\theta =-2$$$$\Rightarrow \theta =-actan2$$$$\mid 1-3i\mid =\sqrt{10}and1-3i=\sqrt{10}(\frac{1}{\sqrt{10}}-\frac{3}{\sqrt{10}}i)=r{e}^{i\phi }$$$$\Rightarrow r=\sqrt{10}andcos\phi =\frac{1}{\sqrt{10}}andsin\phi =-\frac{3}{\sqrt{10}}$$$$\Rightarrow tan\phi =-3\Rightarrow \phi =-arctan3so$$$$Arg\left(\frac{-1+2i}{1-3i}\right)\equiv \theta -\phi \left[2\pi \right]\equiv arctan3-arctan2\left[2\pi \right].$$

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