Finding real values for coefficients

Question

z^2 + az + b = 0

I need to find real values for a and b such that the expression has z=5-14i as a root. I tried to sub in the value for z but obviously got nowhere because there's two unknowns. Is there a trick I'm missing?

Answer 1

If you want $a$ and $b$ to be real numbers, recall that if a polynomial with real coefficients has $z \in \mathbb{C}$ as a root, it also has $\overline{z} \in \mathbb{C}$ as a root. This means the two roots of this polynomial must be $5-14i$ and $5+14i$. Then the polynomial must be:

$$(z-5+14i)(z-5-14i) = z^2 -(5+14i + 5-14i)z +(5-14i)(5+14i)$$ $$= z^2 -10z + 221$$