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.
- Quadratic formula:
$$\begin{align} z_1,z_2 &=-\frac{p}{2}\pm\sqrt{\left( \frac{p}{2}\right)^2-q} \\ & \implies z_1,z_2=\frac{1+3i}{2}\pm\sqrt{\left( \frac{(-1-3i)}{2}\right)^2-(-8-i)}\\ &= \frac{1+3i}{2} \pm\sqrt{\frac{24+10i}{4} }=\frac{1+3i \pm\sqrt{24+10i}}{2}\end{align}$$
- 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?
Best Answer
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 $\pm(5+i).$