If z - x - iy - z -1- z -1- and arg z -1- z -1-4- then z -A- -2-1B- -2-1 i- -2-1D- 1-2

The correct option is C |z+1| = |z 1| ⇒ (x+1)^2 + y^2 = (x 1)^2 + y^2 ⇒ x = 0 Then z 1/z+1 = iy 1/iy+1 = (iy 1)( iy+1)/1+y^2 = y^2 1+2yi/y^2+1 arg z 1/z+1 = π ...

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Byju's Answer

Standard XII

Mathematics

Euler's Representation

If z = x + iy...

Question

If $z$ = $x + i y$ , $| z + 1 |$ = $| z - 1 |$ and arg $⟮ \frac{z - 1}{z + 1} ⟯$ = $\frac{π}{4}$ , then z =

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Solution

The correct option is A

$| z + 1 |$ = $| z - 1 |$

$\Rightarrow (x + 1)^{2} + y^{2}$ = $(x - 1)^{2} + y^{2} \Rightarrow x$ = 0

Then $\frac{z - 1}{z + 1}$ = $\frac{i y - 1}{i y + 1}$ = $\frac{(i y - 1) (- i y + 1)}{1 + y^{2}}$ = $\frac{y^{2} - 1 + 2 y i}{y^{2} + 1}$

arg $⟮ \frac{z - 1}{z + 1} ⟯$ = $\frac{π}{4} \Rightarrow y^{2} - 1$ = $2 y, y > 0$

$\Rightarrow y$ = $1 + \sqrt{2}$

Hence z = $(1 + \sqrt{2}) i$

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If z = x+iy, |z+1| = |z−1| and arg ⟮z−1z+1⟯ = π4, then z =

The correct option is A |z+1| = |z−1| ⇒(x+1)2+y2 = (x−1)2+y2⇒x = 0 Then z−1z+1 = iy−1iy+1 = (iy−1)(−iy+1)1+y2 = y2−1+2yiy2+1 arg ⟮z−1z+1⟯ = π4⇒y2−1 = 2y,y>0 ⇒y = 1+√2 Hence z = (1+√2)i

If $z$ = $x + i y$ , $| z + 1 |$ = $| z - 1 |$ and arg $⟮ \frac{z - 1}{z + 1} ⟯$ = $\frac{π}{4}$ , then z =