Ask Uncle Colin: Complex quadratics with real values

Dear Uncle Colin,

I was wondering: given a quadratic function with real coefficients, what complex arguments lead to real answers?

- Researching Equations And Lines

Hi, REAL, and thanks for your message!

This turns out to be simpler than I expected: if you have a quadratic $f(z) = az^2 + bz + c$ with real coefficients $a$, $b$ and $c$, and $z = x + y\i$ (with $x$ and $y$ real), you can simply find the complex part as $f_i(z) = 2axy + by$.

For $f(z)$ to be real, $f_i(z)$ must be zero, which gives $y(2ax + b)=0$.

If $y=0$, we have a real number to start with (which I think we already knew would give real answers); alternatively, if $x = \frac{-b}{2a}$, we get a real answer. This is neat: it's the line of complex numbers perpendicular to the real axis, passing through the vertex of the (real) graph.

What about complex coefficients?

That leads to a natural extension: what if we don't assert that the coefficients are real?

This gets a bit messy, but let's roll with it: $f_i(z)=a_i (x^2 - y^2) + 2a_r xy + b_i x + b_r y + c_i$. Where is this zero? It looks for all the world like a hyperbola!

However, I'll leave the properties of it as something for you to research once you've finished with equations and lines!

Hope that helps,

- Uncle 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.


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