If z is a complex number such that {z^2} = {( | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c)Let $z = x + iy$, then its conjugate $\overline z = x - iy$
Given that ${z^2} = {(\overline z )^2}$
==> ${x^2} - {y^2} + 2ixy = {x^2} - {y^2}

Vedclass · Accepted Answer

Either $z$ is purely real or purely imaginary

Vedclass · Answer

(c)Let $z = x + iy$, then its conjugate $\overline z = x - iy$
Given that ${z^2} = {(\overline z )^2}$
==> ${x^2} - {y^2} + 2ixy = {x^2} - {y^2} - 2ixy$==> $4ixy = 0$
If $x 
e 0$ then $y = 0$and if $y 
e 0$then $x = 0$
.

If $z$ is a complex number such that ${z^2} = {(\bar z)^2},$ then

Similar Questions

The solutions of equation in $z$, $| z |^2 -(z + \bar{z}) + i(z - \bar{z})$ + $2$ = $0$ are $(i = \sqrt{-1})$

If $z = 3 + 5i,\,\,{\rm{then }}\,{z^3} + \bar z + 198 = $

Find the modulus and argument of the complex numbers:

$\frac{1}{1+i}$

If $z$ is a purely real number such that ${\mathop{\rm Re}\nolimits} (z) < 0$, then $arg(z)$ is equal to

If ${Z_1} \ne 0$ and $Z_2$ be two complex numbers such that $\frac{{{Z_2}}}{{{Z_1}}}$ is a purely imaginary number, then $\left| {\frac{{2{Z_1} + 3{Z_2}}}{{2{Z_1} - 3{Z_2}}}} \right|$ is equal to