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

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.

\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}

1. 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?

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 $$\pm(5+i).$$

• Thanks for explaining that. I am new to complex roots and complex polynomials so I didn't immediately think of doing this. Thank you! – Nullspace Dec 16 '18 at 10:46
• You're welcome :) – user376343 Dec 16 '18 at 10:55

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

• I should have seen that. Thank you! – Nullspace Dec 16 '18 at 10:28

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

• I should have seen that. Thank you! – Nullspace Dec 16 '18 at 10:28

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

• I should have seen that. Thank you! – Nullspace Dec 16 '18 at 10:28

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.

• Ah I see. I would have do use de moivres formula then right? – Nullspace Dec 16 '18 at 10:29
• For odd prime roots yes. For square roots no. – Oscar Lanzi Dec 16 '18 at 10:36
• So for example: $z^8=-3+\sqrt{3}i$ I could solve without de moivres formula? – Nullspace Dec 16 '18 at 10:44
• Yup, you just do three square root extractions. The algebra gets complicated but the concept works. – Oscar Lanzi Dec 16 '18 at 10:57