Question Number 127366 by mohammad17 last updated on 29/Dec/20
$${Appointed}\:{at}\:{the}\:{nodal}\:{level}\:{of}\:{the}\:{group} \\ $$$$ \\ $$$$\mid{z}+\overset{−} {{z}}\mid^{\mathrm{2}} +\left({z}−\overset{−} {{z}}\right)=\mathrm{4}? \\ $$
Commented by JDamian last updated on 29/Dec/20
$${z}=\pm\mathrm{1} \\ $$
Commented by mohammad17 last updated on 29/Dec/20
$${how}\:{sir}\:{can}\:{you}\:{give}\:{me}\:{steps}? \\ $$
Commented by JDamian last updated on 29/Dec/20
$${z}\equiv{a}+{bi} \\ $$$$ \\ $$$${z}+\overset{−} {{z}}=\mathrm{2}{a} \\ $$$${z}−\overset{−} {{z}}=\mathrm{2}{bi} \\ $$$$\mid{z}+\overset{−} {{z}}\mid^{\mathrm{2}} +\left({z}−\overset{−} {{z}}\right)=\mathrm{4} \\ $$$$\mid\mathrm{2}{a}\mid^{\mathrm{2}} +\mathrm{2}{bi}=\mathrm{4}\Rightarrow\begin{cases}{{b}=\mathrm{0}}\\{\mathrm{4}\mid{a}\mid^{\mathrm{2}} =\mathrm{4}\:\Rightarrow\mid{a}\mid^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{a}=\pm\mathrm{1}}\end{cases} \\ $$$$ \\ $$
Commented by mohammad17 last updated on 29/Dec/20
$${thank}\:{you}\:{sir}\:{are}\:{you}\:{can}\:{help}\:{me}\:{in}\:{allater}\:{question}? \\ $$
Answered by liberty last updated on 29/Dec/20
$${z}={a}+{bi}\:\Rightarrow\overset{−} {{z}}={a}−{bi} \\ $$$$\mid{z}+\overset{−} {{z}}\mid^{\mathrm{2}} =\:\mathrm{4}{a}^{\mathrm{2}} \:;\:{z}−\overset{−} {{z}}=\mathrm{2}{bi}\: \\ $$$$\Leftrightarrow\:\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}{bi}\:=\:\mathrm{4}+\mathrm{0}{i}\:\rightarrow\begin{cases}{\mathrm{4}{a}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow{a}=\pm\mathrm{1}}\\{{b}=\mathrm{0}}\end{cases} \\ $$$$\therefore\:{z}\:=\:\pm\:\mathrm{1} \\ $$