How to solve $z^2-(1-3i)z-2i-2=0$

Question

I've tried two ways, but get stuck.

I've tried to simplify, but didn't know what to do next, and I've tried to solve it like a Quadratic equation but got stuck too.

One way got me this:

$$\frac{z}{2} \times (-1+3i+z)-1-i=0$$ - don't know what next.

option b, got me mess, while trying to make the equation into a Quadratic.

Answer 1

I'd define $z = x + yi$ and substitute:

$$(x+yi)^2 - (1 - 3i)(x+yi) - 2i - 2 = 0$$ $$x^2 + 2xyi - y^2 - x - yi + 3xi - 3y^2 - 2i - 2 = 0$$

This gives you two equations (one for the real part and one for the imaginary part) with two unknowns:

$$x^2 - 4y^2 - x - 2 = 0$$ $$3x - y + 2xy - 2 = 0$$

From here you can try some substitutions to remove the coupling, or see if $z = re^{\theta i}$ is easier. In any case, the main trick is that the real part and imaginary must both be equal on both sides.

Can you take it from here?

Answer 2

Using the quadratic formula for a general quadratic $ax^2+bx+c=0$ which is $$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=0$$ for $z^2-(1-3i)z-2i-2=0$ $$z=\frac{1-3i\pm \sqrt{-8-6i+8i+8}}{2}$$ $$\implies z=\frac{1-3i\pm \sqrt{2i}}{2}$$

Answer 3

Just use the standard quadratic formula - it works for complex numbers as well as for reals.

Answer 4

Rational root testing in Gaussian integers

You can also seek a "rational" root. With complex coefficients having integer real and imaginary parts you seek roots of the firm $(a+bi)/(c+di)$ where $a,b,c,d$ are integers.

Here the leading coefficient us $1$ so we must have just $a+bi$. We then have the constant term

$-2-2i=(1)(-2-2i)=(i)(-2+2i)=(2)(-1-i)=(2i)(-1+i),$

and if the roots have integer parts must be one of the above factors or its negative. Since the roots must add up to $1-3i$ there would have to be a root with imaginary part $-2$, so we try

$-2i,\pm2-2i.$

When the candidate root $-2i$ holds, the other root is $(1-3i)-(-2i)=1-i$.