How can I solve complex number equation? $-\frac12·\frac{iz-2z}{z^2}+\frac{-1-i}{2i-2z}=
\frac{\frac{-3}4-\frac12i}{z}$
if $z=a+bi$,
How to find $a$ and $b$?
Thank you.
 A: The procedure is the same as solving a non-complex equation. First one has to set the domain to be $\mathbb{C} \setminus \{0,i\} $, else you can get a division by zero.
Then we multiply the equation with $2z^2(i-z)$ in order to get rid of the fractions. So we get 
\begin{align*}
(2z-iz)(i-z) + (-1-i)z^2 &= -2z(i-z)(\dfrac{3}{4} + \dfrac{1}{2}i)\\
(-2z^2 + 2iz + z +iz^2) + (-z^2-iz^2) &= (z^2-iz)(\dfrac{3}{2} + i)\\
-3z^2+z + 2iz &= \frac{3}{2}z^2 + iz^2 -\frac{3}{2}iz + z \\
\dfrac{9}{2}z^2+iz^2-\dfrac{7}{2}iz &= 0 \\
z^2(\dfrac{9}{2}+i)-z(\dfrac{7}{2}i) &= 0
\end{align*}
We know that the $z$ can't be zero from the solution domain from $z$, so we can divide the equation by z:
\begin{align*}
z(\dfrac{9}{2}+i)-\dfrac{7}{2}i &= 0 \\
z(\dfrac{9}{2}+i)&= \dfrac{7}{2}i \\
z &= \frac{\frac{7}{2}i}{\frac{9}{2} + i}
\end{align*}
This complex division should not be a problem for you and you get your solution.
In the case of a quadratic equation at the end: use the quadratic formula as you would with a non-complex equation. Just pay attention to the complex square-root.
A: $$-\frac12·\frac{iz-2z}{z^2}+\frac{-1-i}{2i-2z}=
\frac{\frac{-3}4-\frac12i}{z}\Longleftrightarrow$$
$$\frac{(1+2i)-3z}{2iz-2z^2}=
\frac{-\frac{i}{2}-\frac{3}{4}}{z}\Longleftrightarrow$$
$$4z((1+2i)-3z)=
(-3-2i)(2iz-2z^2)\Longleftrightarrow$$
$$(4+8i)z-12z^2=(6+4i)z^2+(4-6i)z\Longleftrightarrow$$
$$(-18-4i)z^2+14iz=0\Longleftrightarrow$$
$$-2z((9+2i)z-7i)=0\Longleftrightarrow$$
$$z((9+2i)z-7i)=0\Longleftrightarrow$$
$$(9+2i)z-7i=0\Longleftrightarrow$$
$$ (9+2i)z=7i\Longleftrightarrow$$
$$z=\frac{7i}{9+2i}$$
