Let $z$ be a purely imaginary number such that ${\mathop{\rm Im}\nolimits} (z) < 0$. Then $arg\,(z)$ is equal to

If $z_1 = 1+2i$ and $z_2 = 3+5i$ , then ${\mathop{\rm Re}\nolimits} \,\left( {\frac{{{{\overline Z }_2}{Z_1}}}{{{Z_2}}}} \right) = $

If complex number $z = x + iy$ is taken such that the amplitude of fraction $\frac{{z - 1}}{{z + 1}}$ is always $\frac{\pi }{4}$, then

If $|z_1| = 2 , |z_2| =3 , |z_3| = 4$ and $|2z_1 +3z_2 +4z_3| =9$ ,then value of $|8z_2z_3 +27z_3z_1 +64z_1z_2|$ is equal to:-

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 $z$ )

 Find the conjugate of $\frac{(3-2 i)(2+3 i)}{(1+2 i)(2-i)}$.