Special feature of the function f(z) = $|i + z|^2 + az + 3$

Question

I have to solve following problem: Find all the values of a (a is a real number) that the function f : $f(z) = |i + z|^2 + az + 3$ (z is a complex number, i is an imaginary unit) has a following feature: if $f(u) = 0$ then $f(u') = 0$ (where u' is a complex conjugate of u). Do you have any ideas what is the best approach to solve it?

Answer 1

Since $\vert z + i \vert$ is real and non-negative, so is $\vert z + i \vert^2$; thus, since $a \in \Bbb R$, $f(u) = 0$, for $a \ne 0$, implies the solution $u$ must be real if it exists, so $\bar u = u$; and hence $f(\bar u) = f(u) = 0$. Note there is no solution for $a = 0$ since $\vert z + i \vert^2 + 3 > 0$ for all $z \in \Bbb C$. In fact since $u$ must be real, we have $\vert u + i \vert^2 = (u + i)(u - i) = u^2 + 1$, so $f(u) = 0$ becomes

$u^2 + au + 4 = 0, \tag{1}$

implying

$u = \dfrac{1}{2}(-a \pm \sqrt{a^2 - 16}); \tag{2}$

these $u$ are real when

$\vert a \vert \ge 4; \tag{3}$

furthermore, working backwards from (2), we have,

$2u + a = \pm \sqrt{a^2 - 16}, \tag{4}$

whence

$4u^2 + 4au + a^2= a^2 - 16, \tag{5}$

or

$4u^2 + 4au + 16 = 0, \tag{6}$

or, upon division by $4$͵

$u^2 + au+ 4 = 0, \tag{7}$

that is,

$0 = (u^2 + 1) + au + 3 = (u - i)(u + i) + au + 3$ $= \vert u + i \vert^2 + au + 3 = f(u).\tag{8}$

(8) shows that, with (3) holding, we can find a real $u$ such that $f(u) = 0$, hence $f(\bar u) = f(u) = 0$ as well.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!