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