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How does one solve expressions such as $(1-i)z^2-2iz-4=0$

Own attempt

$$\begin{align} &z^2-\frac{2iz}{1-i}-\frac{4}{1-i}=z^2-z(1-i)-(2+2i)=0\\\iff&z^2-z(1-i)-\frac{i}{2} = 2+\frac{3i}{2} \iff (z+(\frac{1}{2}-\frac{i}{2}))^2=2+\frac{3i}{2}\\\iff &z+(\frac{1}{2}-\frac{i}{2}) = \pm\sqrt{2+\frac{3i}{2}}\end{align}$$

If my method was a slow choice, feel free to use a different method.

Edit 1. Approached in different way according to comments

$$\begin{align} &2z^2-z(2i-2) - (4+4i) = z^2-z(i-1)-(2+2i) = 0\\&z = \frac{i-1}{2}\pm\sqrt{\frac{(i-1)^2}{4}+(2+2i)}\\&z=\frac{i-1}{2}\pm\sqrt{\frac{-1-2i+1}{4}+(2+2i)}\\&z= \frac{i-1}{2}\pm\sqrt{2+\frac{3i}{2}} \end{align}$$

However, I still can't figure the rest out. I was thinking about using de Moivre's formula by converting the complex number inside the square root sign into polar form, however, it's not going to help me, I figure.

I want to get to the solutions $z_1 = (1+i)$ and $z_2 = -2$

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  • $\begingroup$ Quadratic formula works too. $\endgroup$ – Antonio Vargas Oct 21 '14 at 22:37
  • $\begingroup$ To avoid quotients, you can instead multiply through by $1+i$ $\endgroup$ – Semiclassical Oct 21 '14 at 22:39
  • $\begingroup$ I tried multiplying through by $(1+i)$, however, I still don't know how to find the two sought solutions. $\endgroup$ – B. Lee Oct 21 '14 at 23:04
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I found that if you study the patterns between roots and coefficients, one finds that the roots $x_1$ and $x_2$ are:

$$\begin{cases}x_1+x_2 = -a_1/a_2\\x_1\cdot x_2 = a_0/a_2\end{cases}$$ Where $a_i$ is the coefficient in front of the $i'th$ term. This means that if you study the given equation, one finds that $$x_1+x_2 = 2i/(1-i)\\x_1\cdot x_2 = -4/(1-i)$$

Which solves to $x_1 = -2$ and $x_2 = (1+i)$

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Leading on from your working, consider $z=x+iy= \sqrt{2+ \frac{3i}{2}}$. By solving for x and y, you can effectively square root $2+\frac{3i}{2}$.

With your simultaneous equations, you might like to consider $|z|^2$ for a quick solution to the simultaneous equations.

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