Question Number 85668 by Rio Michael last updated on 23/Mar/20
$$\mathrm{z}\:=\:\mathrm{2}\:+\:\mathrm{i}\: \\ $$$$\mathrm{find}\:\mathrm{arg}\left(\mathrm{z}\right) \\ $$
Commented by mathmax by abdo last updated on 24/Mar/20
![∣z∣=(√(2^2 +1^2 ))=(√5) ⇒ z =(√5){(2/(√5))+(1/(√5))i} =r e^(iθ) ⇒ r=(√5)and cosθ (2/(√5)) ,sinθ =(1/(√5)) ⇒tanθ =(1/2) ⇒θ= arctan((1/2)) =(π/2)−arctan(2) arg(z)≡(π/2) −arctan(2) [2π]](Q85673.png)
$$\mid{z}\mid=\sqrt{\mathrm{2}^{\mathrm{2}} \:+\mathrm{1}^{\mathrm{2}} }=\sqrt{\mathrm{5}}\:\Rightarrow\:{z}\:=\sqrt{\mathrm{5}}\left\{\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}+\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}{i}\right\}\:={r}\:{e}^{{i}\theta} \:\Rightarrow \\ $$$${r}=\sqrt{\mathrm{5}}{and}\:{cos}\theta\:\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\:\:,{sin}\theta\:=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\:\Rightarrow{tan}\theta\:=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\theta=\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{2}\right) \\ $$$${arg}\left({z}\right)\equiv\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\mathrm{2}\right)\:\left[\mathrm{2}\pi\right] \\ $$
Commented by Rio Michael last updated on 24/Mar/20
$$\mathrm{thanks}\:\mathrm{sir} \\ $$