The answer is “no”. Note that \begin{eqnarray} & & z^2 + (-2+2i)z - (1+2i) = 0 \\ & \Leftrightarrow & z^2 + (-2+2i)z = 1+2i \\ & \Leftrightarrow & z^2 + (-2+2i)z + \left (\frac{-2+2i}{2} \right)^2 = 1 + 2i + \left(\frac{-2+2i}{2}\right)^2 \\ & \Leftrightarrow & \left (z + \frac{-2+2i}{2} \right)^2 = 1 + 2i + (-1 +i)^2 \\ & \Leftrightarrow & (z + (-1+i))^2 = 1+2i - 2i \\ & \Leftrightarrow & (z + (-1+i))^2 = 1 \\ \end{eqnarray} Since the square roots of \(1\) are \(1\) and \(-1\), we must have \(z = 1 - (-1+i) = 2-i\) or \(z = -1 - (-1 +i) = -i\). In summary, the solutions are \(2-i\) and \(-i\) and neither is a real number.