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Question Number 21109 by Tinkutara last updated on 13/Sep/17

STATEMENT-1 : The locus of z, if arg(((z − 1)/(z + 1))) = (π/2) is a circle. and STATEMENT-2 : ∣((z − 2)/(z + 2))∣ = (π/2), then the locus of z is a circle.

$$\mathrm{STATEMENT}-\mathrm{1}\::\:\mathrm{The}\:\mathrm{locus}\:\mathrm{of}\:{z},\:\mathrm{if} \\ $$$$\mathrm{arg}\left(\frac{{z}\:−\:\mathrm{1}}{{z}\:+\:\mathrm{1}}\right)\:=\:\frac{\pi}{\mathrm{2}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\boldsymbol{\mathrm{and}} \\ $$$$\mathrm{STATEMENT}-\mathrm{2}\::\:\mid\frac{{z}\:−\:\mathrm{2}}{{z}\:+\:\mathrm{2}}\mid\:=\:\frac{\pi}{\mathrm{2}},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\:{z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}. \\ $$

Commented by youssoufab last updated on 13/Sep/17

arg(((z−1)/(z+1)))=(π/2) ⇔arg(((z−1)/(z+1)))=arg(i) ⇔((z−1)/(z+1))≡i[2π] ⇔z−1≡(z+1)i[2π] ⇔z−1≡zi+i[2π] ⇔z(1−i)≡i+1[2π] ⇔z≡((i+1)/(1−i))[2π] ⇔z≡(((i+1)^2 )/((1−i)(i+1)))[2π] ⇔z≡((−1+2i+1)/(1+1))[2π] ⇔z≡i[2π] z is a cercle⇒z=i

$$\mathrm{arg}\left(\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{arg}\left(\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\right)=\mathrm{arg}\left({i}\right) \\ $$$$\Leftrightarrow\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\equiv{i}\left[\mathrm{2}\pi\right] \\ $$$$\Leftrightarrow{z}−\mathrm{1}\equiv\left({z}+\mathrm{1}\right){i}\left[\mathrm{2}\pi\right] \\ $$$$\Leftrightarrow{z}−\mathrm{1}\equiv{zi}+{i}\left[\mathrm{2}\pi\right] \\ $$$$\Leftrightarrow{z}\left(\mathrm{1}−{i}\right)\equiv{i}+\mathrm{1}\left[\mathrm{2}\pi\right] \\ $$$$\Leftrightarrow{z}\equiv\frac{{i}+\mathrm{1}}{\mathrm{1}−{i}}\left[\mathrm{2}\pi\right] \\ $$$$\Leftrightarrow{z}\equiv\frac{\left({i}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{i}\right)\left({i}+\mathrm{1}\right)}\left[\mathrm{2}\pi\right] \\ $$$$\Leftrightarrow{z}\equiv\frac{−\mathrm{1}+\mathrm{2}{i}+\mathrm{1}}{\mathrm{1}+\mathrm{1}}\left[\mathrm{2}\pi\right] \\ $$$$\Leftrightarrow{z}\equiv{i}\left[\mathrm{2}\pi\right] \\ $$$${z}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cercle}\Rightarrow{z}={i} \\ $$$$ \\ $$

Commented by Tinkutara last updated on 13/Sep/17

Locus of z will not be a complete circle, but why it is not, is my doubt.

$$\mathrm{Locus}\:\mathrm{of}\:{z}\:\mathrm{will}\:\mathrm{not}\:\mathrm{be}\:\mathrm{a}\:\mathrm{complete}\:\mathrm{circle}, \\ $$$$\mathrm{but}\:\mathrm{why}\:\mathrm{it}\:\mathrm{is}\:\mathrm{not},\:\mathrm{is}\:\mathrm{my}\:\mathrm{doubt}. \\ $$

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