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