Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.
We have, $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$
$=\frac{6+9 i-4 i+6}{2-i+4 i+2}=\frac{12+5 i}{4+3 i} \times \frac{4-3 i}{4-3 i} $
$=\frac{48-36 i+20 i+15}{16+9}=\frac{63-16 i}{25}=\frac{63}{25}-\frac{16}{25} i$
Therefore, conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$ is $\frac{63}{25}+\frac{16}{25} i$.
If ${z_1}$ and ${z_2}$ are two complex numbers satisfying the equation $\left| \frac{z_1 +z_2}{z_1 - z_2} \right|=1$, then $\frac{{{z_1}}}{{{z_2}}}$ is a number which is
Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$
Let $z_1$ and $z_2$ be any two non-zero complex numbers such that $3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$. If $z = \frac{{3{z_1}}}{{2{z_2}}} + \frac{{2{z_2}}}{{3{z_1}}}$ then
The solution of the equation $|z| - z = 1 + 2i$ is
If $Arg(z)$ denotes principal argument of a complex number $z$, then the value of expression $Arg\left( { - i{e^{i\frac{\pi }{9}}}.{z^2}} \right) + 2Arg\left( {2i{e^{-i\frac{\pi }{{18}}}}.\overline z } \right)$ is