Let the complex number $z = x + iy$ be such that $\frac{2z - 3i}{2z + i}$ is purely imaginary. If $x + y^2 = 0$,then $y^4 + y^2 - y$ is equal to:

If $x = -2 + \sqrt{-3}$,then the value of $2x^4 + 5x^3 + 7x^2 - x + 38$ is equal to

If the real part of the complex number $z = \frac{3 + 2i \cos \theta}{1 - 3i \cos \theta}$, where $\theta \in (0, \frac{\pi}{2})$, is zero, then the value of $\sin^2 3\theta + \cos^2 \theta$ is equal to:

Let the complex numbers $\alpha$ and $\frac{1}{\bar{\alpha}}$ lie on the circles $|z-z_0|^2=4$ and $|z-z_0|^2=16$ respectively,where $z_0=1+i$. Then,the value of $100|\alpha|^2$ is.

If $z_1=1-2 i$,$z_2=1+i$,and $z_3=3+4 i$,then $\left(\frac{1}{z_1}+\frac{3}{z_2}\right) \frac{z_3}{z_2}=$

The biquadratic equation,two of whose roots are $1+i$ and $1-\sqrt{2}$,is