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Question Number 19740 by Tinkutara last updated on 15/Aug/17

If ∣z^2  − 1∣ = ∣z∣^2  + 1, then z lies on

$$\mathrm{If}\:\mid{z}^{\mathrm{2}} \:−\:\mathrm{1}\mid\:=\:\mid{z}\mid^{\mathrm{2}} \:+\:\mathrm{1},\:\mathrm{then}\:{z}\:\mathrm{lies}\:\mathrm{on} \\ $$

Answered by ajfour last updated on 16/Aug/17

 ∣(x+iy)^2 −1∣=x^2 +y^2 +1  ⇒  (x^2 −1−y^2 )^2 +4x^2 y^2 =(x^2 +1+y^2 )^2   ⇒  4x^2 y^2 =2x^2 (2y^2 +2)         x^2 (y^2 +1−y^2 )=0  ⇒  x=0       z lies on the imaginary  axis .

$$\:\mid\left(\mathrm{x}+\mathrm{iy}\right)^{\mathrm{2}} −\mathrm{1}\mid=\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{1} \\ $$$$\Rightarrow\:\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} =\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{4x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} =\mathrm{2x}^{\mathrm{2}} \left(\mathrm{2y}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}−\mathrm{y}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{x}=\mathrm{0} \\ $$$$\:\:\:\:\:\mathrm{z}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{imaginary}\:\:\mathrm{axis}\:. \\ $$

Commented by Tinkutara last updated on 15/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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