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STATEMENT-1-The-locus-of-z-if-arg-z-1-z-1-pi-2-is-a-circle-and-STATEMENT-2-z-2-z-2-pi-2-then-the-locus-of-z-is-a-circle-

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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.
STATEMENT−1:Thelocusofz,ifarg(z−1z+1)=π2isacircle.andSTATEMENT−2:∣z−2z+2∣=π2,thenthelocusofzisacircle.
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
arg(z−1z+1)=π2⇔arg(z−1z+1)=arg(i)⇔z−1z+1≡i[2π]⇔z−1≡(z+1)i[2π]⇔z−1≡zi+i[2π]⇔z(1−i)≡i+1[2π]⇔z≡i+11−i[2π]⇔z≡(i+1)2(1−i)(i+1)[2π]⇔z≡−1+2i+11+1[2π]⇔z≡i[2π]zisacercle⇒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.
Locusofzwillnotbeacompletecircle,butwhyitisnot,ismydoubt.

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