Complex Numbers, Finding solutions to $z$

Question

Can anyone help me find the complex solutions to $z^2-2(1+i)z+5i=0$ using the quadratic formula? I get stuck with simplifying the equation.

Answer 1

Rather than use the quadratic formula, you may find it easier to complete the square. We have that

$$z^2-2(1+i)z=-5i.$$

Completing the square:

$$z^2-2(1+i)z+2i=-5i+2i.$$

Hence,

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

At this point, we need to find a complex number $w$ with $w^2=-3i$. This is easier if we use the polar form. We have that

$$-3i=3\cdot(-i)=3e^{i\frac{3\pi}{2}}$$

Hence

$$\begin{align*} w &= \pm\sqrt{3}e^{i\frac{3\pi}{4}} \\ &=\pm\sqrt{3}\cdot\left(\cos\frac{3\pi}{4}+i\cdot\sin\frac{3\pi}{4}\right) \\ &=\pm\sqrt{3}\cdot\left(-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right) \\ &=\frac{\sqrt{3}}{\sqrt{2}}\cdot(\mp1\pm i) \end{align*}$$

Since we already had that

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

it follows that

$$z-(1+i)=\frac{\sqrt{3}}{\sqrt{2}}\cdot(\mp1\pm i).$$

So that

$$z=\left(1\mp\frac{\sqrt{3}}{\sqrt{2}}\right)+i\cdot\left(1\pm\frac{\sqrt{3}}{\sqrt{2}}\right).$$

Answer 2

OP's way

$$z^2-2(1+i)z+5i=0 \implies z=\frac{2(1+i) \pm \sqrt{(4(1+i)^2-20i}}{2}.$$$$\implies z=\frac{2(1+i)\pm 2\sqrt{3} \sqrt{-i}}{2}=z=\frac{2(1+i)\pm 2\sqrt{3} (1-i)/\sqrt{2}}{2}.$$ $$\implies z= \left(1\pm \sqrt{\frac{3}{2}}\right)+i\left(1\mp\sqrt{\frac{3}{2}}\right).$$