Question Number 207563 by hardmath last updated on 18/May/24
$$\mathrm{z}\:\:+\:\:\mid\mathrm{z}\mid\:\:=\:\:\mathrm{1}\:\:+\:\:\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\varphi}\:=\:? \\ $$
Answered by Frix last updated on 18/May/24
$${z}+\mid{z}\mid=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i} \\ $$$$\mid{z}\mid=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}−{z} \\ $$$$\mid{z}\mid\in\mathbb{R}\:\Rightarrow\:{z}={a}+\sqrt{\mathrm{3}}\mathrm{i} \\ $$$$\mid{a}+\sqrt{\mathrm{3}}\mathrm{i}\mid=\mathrm{1}−{a} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}=\mathrm{1}−{a}\:\Rightarrow\:{a}=−\mathrm{1} \\ $$$${z}=−\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{2e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow\:\varphi=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by hardmath last updated on 19/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Answered by A5T last updated on 18/May/24
$${z}={a}+{bi}\Rightarrow\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${z}+\mid{z}\mid=\left({a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)+{bi}=\mathrm{1}+{i}\sqrt{\mathrm{3}}\Rightarrow{b}=\sqrt{\mathrm{3}} \\ $$$${a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1}\Rightarrow{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}=\mathrm{1}\Rightarrow{a}^{\mathrm{2}} +\mathrm{3}=\mathrm{1}+{a}^{\mathrm{2}} −\mathrm{2}{a} \\ $$$$\Rightarrow{a}=−\mathrm{1}\Rightarrow{z}=−\mathrm{1}+{i}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{z}=\mathrm{2}{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$