Solve: $i x^2 - 3 x - 2 i = 0$

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 $i=\sqrt{-1}$,then $\sum_{n=0}^{\infty}\left(\frac{i}{3}\right)^n=$

If $a = \cos (2\pi /7) + i\sin (2\pi /7)$,then the quadratic equation whose roots are $\alpha = a + a^2 + a^4$ and $\beta = a^3 + a^5 + a^6$ is

If $e^{it} = \cos t + i \sin t$ and $e^{-it} = \cos t - i \sin t$, then $\cosh(x + iy) - \cosh(x - iy) =$

Find the complex number $z$ satisfying the equations $\left| \frac{z - 12}{z - 8i} \right| = \frac{5}{3}$ and $\left| \frac{z - 4}{z - 8} \right| = 1$.