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

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4-1.Complex numbers

medium

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

$z$ is purely real

$z$ is purely imaginary

Either $z$ is purely real or purely imaginary

None of these

Solution

(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 $x \ne 0$ then $y = 0$and if $y \ne 0$then $x = 0$
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Standard 11

Mathematics

$(z + a)(\bar z + a)$, where $a$ is real, is equivalent to

easy

Let $z$ and $w$ be the two non-zero complex numbers such that $|z|\, = \,|w|$ and $arg\,z + arg\,w = \pi $. Then $z$ is equal to

medium

(IIT-1995)

If $z$ is a complex number such that $| z | = 4$ and $arg \,(z) = \frac {5\pi }{6}$ , then $z$ is equal to

normal

If $z$ and $\omega$ are two complex numbers such that $|z \omega|=1$ and $\arg (z)-\arg (\omega)=\frac{3 \pi}{2}$, then $\arg \left(\frac{1-2 \bar{z} \omega}{1+3 \bar{z} \omega}\right)$ is:

(Here arg(z) denotes the principal argument of complex number $z$ )

hard

(JEE MAIN-2021)

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hard

(JEE MAIN-2013)