Find all the solutions of $z^2-(1+3i)z-8-i=0$

Question

I am stuck on a problem and I was hoping someone could tell me what I am doing wrong.

I want to find all the roots of:

$$z^2-(1+3i)z-8-i=0$$

There are two ways I tried to approach this.

1. Quadratic formula:

$$\begin{align} z_1,z_2 &=-\frac{p}{2}\pm\sqrt{\left( \frac{p}{2}\right)^2-q} \\ & \implies z_1,z_2=\frac{1+3i}{2}\pm\sqrt{\left( \frac{(-1-3i)}{2}\right)^2-(-8-i)}\\ &= \frac{1+3i}{2} \pm\sqrt{\frac{24+10i}{4} }=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{align}$$

2. Completing the square:

$$\begin{aligned} z^2-(1+3i)z-8-i &=0 \\ & \iff \left(z-\left( \frac{1+3i}{2}\right) \right)^2-8-i=\left( \frac{1+3i}{2}\right)^2 \\ & \iff u^2=\frac{24+10i}{4} \\ & \iff u=\pm\frac{\sqrt{24+10i}}{2} \\ & \iff z_{1,2}=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{aligned} $$

Using both methods I arrive at the same result. However, wolframalpha tells me the roots are:

$$z_1=-2+i \\ z_2=3+2i$$

I have tried everything (writing it in in exponential form, graphing, etc.) but I have no idea how I can arrive at those two roots. What am I doing wrong?

Answer 1

To find square roots $\;\pm(a+ ib)\;$ of $\;24+10i,\;$ solve the equation $$(a+ib)^2=24+10i.$$ Real and imaginary parts of RHS and LHS are equal, and also absolute values:

$$\begin{aligned}a^2-b^2&=24\\ 2ab&=10\\a^2+b^2&=26\end{aligned}$$ We get $a^2=25$ and $ab=5$ hence $b$ has the sign of $a.$ This gives the two square roots $\pm(5+i).$

Answer 2

That's just because the square roots of $24+10i$ are $\pm(5+i)$.

Answer 3

Hint: Use the fact: $$ 24+10i = (5+i)^2$$

Answer 4

Hint: Expanding $$(5+i)^2=25+10i-1$$

Answer 5

See the answer to this question for how to extract complex square roots. Note carefully: when you have an odd prime root of a complex number you can't extract it except with trigonometry, but square roots can be rendered by algebra alone.