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Question Number 85668 by Rio Michael last updated on 23/Mar/20

$$\mathrm{z}=2+\mathrm{i}$$$$\mathrm{find}\arg \left(\mathrm{z}\right)$$

Commented by mathmax by abdo last updated on 24/Mar/20

$$\mid z\mid =\sqrt{2^2+1^2}=\sqrt{5}\Rightarrow z=\sqrt{5}\{\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}i\}=r{e}^{i\theta }\Rightarrow$$$$r=\sqrt{5}andcos\theta \frac{2}{\sqrt{5}},sin\theta =\frac{1}{\sqrt{5}}\Rightarrow tan\theta =\frac{1}{2}\Rightarrow \theta =arctan\left(\frac{1}{2}\right)=\frac{\pi }{2}-arctan\left(2\right)$$$$arg\left(z\right)\equiv \frac{\pi }{2}-arctan\left(2\right)\left[2\pi \right]$$

Commented by Rio Michael last updated on 24/Mar/20

$$\mathrm{thanks}\mathrm{sir}$$

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