Solve the equations $z^2 + (2 - 2i)z + 2i = 0 $ by completing the square

Question

I tried solving this thing by completing the square and I always end up with something like this $(z^2 + (2 - 2i)z - 2i) + 2i + 2i = 0 $ and it doesn't seem like to me that you can factor the part in brackets into a square, any help is appreciated

Solve the equations $z^2 + (2 - 2i)z + 2i = 0 $ by completing the square

Answer 1

Note that $(1-i)^2=-2i$, so $$z^2-(2-2i)z-2i=(z-(1-i))^2.$$

Answer 2

What you want is something like $$ z^2 + (2 - 2i) z + 2i = (z + a + bi)^2 + [\text{some number}] $$ The important thing is getting the $z$ term to match up. In particular, since $(z + a + bi)^2 = z^2 + (2a + 2bi) z + (a + bi)^2$, you want $$ 2a + 2bi = 2 - 2i $$ So $a = 1$ and $b = -1$. That means $a + bi = 1 -i$, so $$ z^2 + (2 - 2i)z + 2i = (z + 1 - i)^2 + \cdots $$ Can you take it from there?