Vi får med $z = x + iy$ att

\left\{ \begin{array}{l} \operatorname{Re} z = 2 \operatorname{Im} z \\ |z - 1 + i| = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} x = 2y \\ |x + iy - 1 + i| = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{c} x = 2y \\ \sqrt{(x - 1)^2 + (y + 1)^2} = 2 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2y \\ ( x - 1)^2 + (y + 1)^2 = 4 \end{array} \right.

vilket ger

\begin{array}{l} (2y - 1)^2 + (y + 1)^2 = 4 \Leftrightarrow \dots \Leftrightarrow 5y^2 - 2y - 2 = 0 \\ \Leftrightarrow \dots \Leftrightarrow y = \frac{1 - \sqrt{11}}{5}, y = \frac{1 + \sqrt{11}}{5}. \end{array}

Detta ger då att $y = \frac{1 - \sqrt{11}}{5}$ ger $x = \frac{2 - 2\sqrt{11}}{5}$ och $y = \frac{1 + \sqrt{11}}{5}$ ger $x = \frac{2 + 2\sqrt{11}}{5}$

vilket ger de komplexa talen $z = \frac{2 - 2\sqrt{11}}{5} + i\frac{1 - \sqrt{11}}{5}$ och $z = \frac{2 + 2\sqrt{11}}{5} + i\frac{1 + \sqrt{11}}{5}$ .