Let-z-be-a-complex-number-If-z-1-z-1-and-arg-z-1-z-1-pi-4-Then-z-is-

Question Number 139118 by EnterUsername last updated on 22/Apr/21

Let z be a complex number. If ∣z+1∣=∣z−1∣ and arg(((z−1)/(z+1)))=(π/4). Then z is ?

$$\mathrm{Let}\:{z}\:\mathrm{be}\:\mathrm{a}\:\mathrm{complex}\:\mathrm{number}.\:\mathrm{If}\:\mid{z}+\mathrm{1}\mid=\mid{z}−\mathrm{1}\mid \\ $$$$\mathrm{and}\:\mathrm{arg}\left(\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\right)=\frac{\pi}{\mathrm{4}}.\:\mathrm{Then}\:{z}\:\mathrm{is}\:?\: \\ $$

Answered by qaz last updated on 22/Apr/21

∣z+1∣=∣z−1∣ ⇒z=a+bi=bi arg(((bi−1)/(bi+1)))=arg((((bi−1)^2 )/(1−b^2 )))=arg(((1−b^2 −2ib)/(1−b^2 ))) =tan^(−1) ((−2b)/(1−b^2 ))=(π/4) 1−b^2 =−2b ⇒b=1±(√2) ⇒z=(1±(√2))i

$$\mid{z}+\mathrm{1}\mid=\mid{z}−\mathrm{1}\mid \\ $$$$\Rightarrow{z}={a}+{bi}={bi} \\ $$$${arg}\left(\frac{{bi}−\mathrm{1}}{{bi}+\mathrm{1}}\right)={arg}\left(\frac{\left({bi}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}−{b}^{\mathrm{2}} }\right)={arg}\left(\frac{\mathrm{1}−{b}^{\mathrm{2}} −\mathrm{2}{ib}}{\mathrm{1}−{b}^{\mathrm{2}} }\right) \\ $$$$=\mathrm{tan}^{−\mathrm{1}} \frac{−\mathrm{2}{b}}{\mathrm{1}−{b}^{\mathrm{2}} }=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{1}−{b}^{\mathrm{2}} =−\mathrm{2}{b} \\ $$$$\Rightarrow{b}=\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{z}=\left(\mathrm{1}\pm\sqrt{\mathrm{2}}\right){i} \\ $$

Commented by mr W last updated on 22/Apr/21

i think only z=(1+(√2))i is solution. because with z=(1−(√2))i: arg (((z−1)/(z+1)))=((3π)/4)≠(π/4)

$${i}\:{think}\:{only}\:{z}=\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){i}\:{is}\:{solution}. \\ $$$${because}\:{with}\:{z}=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){i}: \\ $$$${arg}\:\left(\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\right)=\frac{\mathrm{3}\pi}{\mathrm{4}}\neq\frac{\pi}{\mathrm{4}} \\ $$

Commented by EnterUsername last updated on 22/Apr/21

$${Thanks}\:{Sirs} \\ $$

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Let-z-be-a-complex-number-If-z-1-z-1-and-arg-z-1-z-1-pi-4-Then-z-is-

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