Well, I'd note that you probably made a typo as \(\displaystyle \sqrt{4} = 2\), but in any case, if you're ever unsure of an answer, you can always check it yourself. Let's plug in \(\displaystyle z = i \pm \dfrac{\sqrt4}{2}\) and see what happens:
\(\displaystyle z^2 - 2iz - (1 + i)\)
\(\displaystyle = \left(i + \dfrac{\sqrt{4}}{2} \right)^2 - 2i\left(i + \dfrac{\sqrt{4}}{2} \right) - (1 + i)\)
\(\displaystyle = 2i - \left(-2 + 2i \right) - (1 + i)\)
\(\displaystyle = 2 - 2i\)
Hmm... that's not zero. My best guess as to what you really meant was \(\displaystyle z = i \pm \dfrac{\sqrt{4i}}{2} = i \pm \sqrt{i}\). Plugging that in reveals:
\(\displaystyle z^2 - 2iz - (1 + i)\)
\(\displaystyle \left(i + \sqrt{i} \right)^2 - 2i \left(i + \sqrt{i} \right) - (1 + i)\)
\(\displaystyle \left(-1 - \sqrt{2} + i[1 + \sqrt{2}] \right) - \left( -2 - [1 - i] \cdot \sqrt{2} \right) - (1+ i)\)
\(\displaystyle = 0\)
Okay, that's zero. Substituting in the solution \(\displaystyle z=i - \sqrt{i}\) also gives zero. So, what's the problem? Any particular reason your answer "doesn't seem right?"
