The complex number Z satisfying the equation - z - i - z -1-1 isA- 0 B- 1- iC- -1- iD- 1- i

Let z=x+iy |z i|=|z+1|=1 |z i|=|z+1| ⇒ x^2+(y 1)^2=(x+1)^2+y^2 ⇒ x^2+y^2 2y+1=x^2+2x+1+y^2 ⇒ x= y ⋯(1) |z+1|=1 ⇒ (x+1)^2+x^2=1 [From (1)] ⇒ 2x^2+2x=0 ...

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Standard XII

Mathematics

Distance Formula Using Complex Numbers

The complex n...

Question

The complex number $z$ satisfying the equation $| z - i | = | z + 1 | = 1$ is

0

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1 + i

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- 1 + i

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1 - i

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Solution

The correct options are
A $0$
C $- 1 + i$
Let $z = x + i y$
$| z - i | = | z + 1 | = 1$

$| z - i | = | z + 1 |$
$\Rightarrow x^{2} + (y - 1)^{2} = (x + 1)^{2} + y^{2}$
$\Rightarrow x^{2} + y^{2} - 2 y + 1 = x^{2} + 2 x + 1 + y^{2} \Rightarrow x = - y \dots (1)$

$| z + 1 | = 1$
$\Rightarrow (x + 1)^{2} + x^{2} = 1 [From (1)] \Rightarrow 2 x^{2} + 2 x = 0 \Rightarrow x (x + 1) = 0$
$\Rightarrow x = 0, - 1$
$\Rightarrow y = 0, 1$

$∴ z = 0, - 1 + i$

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The complex number z satisfying the equation |z−i|=|z+1|=1 is

The correct options are A 0 C −1+iLet z=x+iy |z−i|=|z+1|=1 |z−i|=|z+1| ⇒x2+(y−1)2=(x+1)2+y2 ⇒x2+y2−2y+1=x2+2x+1+y2⇒x=−y ⋯(1) |z+1|=1 ⇒(x+1)2+x2=1 [From (1)]⇒2x2+2x=0⇒x(x+1)=0 ⇒x=0,−1 ⇒y=0,1 ∴z=0,−1+i

The complex number $z$ satisfying the equation $| z - i | = | z + 1 | = 1$ is