1.

For any complex number z if |z+1| = |z-1| then show that Re(z) = 0 do it fast please​

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 + iyz=x+iy

Substitute z in the equation,

|x + iy - 1| = |x + iy + 1|∣x+iy−1∣=∣x+iy+1∣

(x + iy - 1) = ± (x + iy + 1)(x+iy−1)=±(x+iy+1)

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

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

x + iy - 1 = - x - iy + 1x+iy−1=−x−iy+1

2x + 2iy = 02x+2iy=0

x + iy = 0x+iy=0

So,

Re(z)=0

And ALSO, IMAGINARY PART of z=0.


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