Ireland_2017

Suppose $v^2+4u=0$, and let $\tau =v/2$. Then, $\tau$ is real and $$r^2-2ir=\frac{v^2}{4}-iv=-u-iv=-w.$$ Thus, the quadratic has a real root.

Conversely, if $\tau$ is a real root of $x^2-2ix+w$, then it is also a real root of $x^2+2ix+\bar{w}$. In other words, $\tau$ satisfies the equations $$r^2-2ir+w=0,r^2+2ir+\bar{w}=0,$$ whence, as $w=u+iv$ and so $w+\bar{w}=2u$ and $w-\bar{w}=2iv$, $$2r^2+2u=0,-4ir+2iv=0.$$ Therefore, $v^2=4r^2=-4u$. Hence, the result.