Question

if zis a complex number such that | z - 1 | = | z + 1 | , show that Re (z) =0 ?​

Answer 1

Answer:

Answer:

• Re(z)=0

Given:

• Z is a complex number.
• And the equation is ,|z-1|=|z+1|

Solution:

As we know that,

$z = x + iy$

Substitute z in the equation,

$|x + iy - 1| = |x + iy + 1|$

$(x + iy - 1) = </strong><strong>±</strong><strong> (x + iy + 1)$

(Here,'+' doesn't satisfy the equation.)

$(x + iy - 1) = - (x + iy + 1)$

$x + iy - 1 = - x - iy + 1$

$2x + 2iy = 0$

$x + iy = 0$

So,

Re(z)=0

And also, imaginary part of z=0.

Answer 2

Answer:

It is proved by taking the magnitude of real part and solving it.

Step-by-step explanation:

Take a complex number, such that $z=x+iy$

$Ix+iy-1I=Ix+i(y-1)I$

$I(x-1)+xyI=Ix+iy+1I$

Taking modulus

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

Squaring both sides

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

$x^{2} +1-2x=x^{2} +1+2x$

$4x=0$

$x=0$

Hence, real part is zero.

$Re(z)=0$

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