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
$$r^2 - 2ir + w = 0, \quad r^2 + 2ir + \bar{w} = 0,$$
whence, as $w = u + iv$ and so $w + \bar{w} = 2u$ and $w - \bar{w} = 2iv$,
$$2r^2 + 2u = 0, \quad -4ir + 2iv = 0.$$
Therefore, $v^2 = 4r^2 = -4u$. Hence, the result.