Dear Uncle Colin,

I'm told that $z=i$ is a solution to the complex quadratic $z^2 + wz + (1+i)=0$, and need to find $w$. I've tried the quadratic formula and completing the square, but neither of those seem to work! How do I solve it?

- Don't Even Start Contemplating A Robust Trial & Error Solution

Hello, DESCART&ES, and thank you for your message!

You'll kick yourself. You're told that when $z=i$, the equation holds, so your starting point should be substituting $z=i$ into the equation and seeing what happens.

You get $-1 + wi + 1 + i = 0$, so $w=-1$.

For the record, the quadratic formula should work (although it's WAY overkill): $z = \frac{-w \pm \sqrt{w^2 - 4(1+i)}}{2}$. For one of those roots to be equal to $i$, you have: $-w \pm \sqrt{w^2 - 4 - 4i} = 2i$, or $\pm \sqrt{w^2 - 4 - 4i} = w + 2i$.

Squaring both sides gives $w^2 - 4 - 4i = w^2 + 4wi - 4$, so $-4i = 4wi$ and again, $w=-1$.

Hope that helps!

-- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.