Aus Online Mathematik Brückenkurs 2

Even if the equation contains complex numbers as coefficients, we treat is as an ordinary second-degree equation and solve it by completing the square taking the root.

We complete the square on the left-hand side:

\displaystyle \begin{align} & \left( z-\left( 1+i \right) \right)^{2}-\left( 1+i \right)^{2}+2i-1=0 \\ & \left( z-\left( 1+i \right) \right)^{2}-\left( 1+2i+i^{2} \right)+2i-1=0 \\ & \left( z-\left( 1+i \right) \right)^{2}-1-2i+1+2i-1=0 \\ & \left( z-\left( 1+i \right) \right)^{2}-1=0 \\ \end{align}

Now, we see that the equation has the solutions

\displaystyle z-\left( 1+i \right)=\pm 1\quad \Leftrightarrow \quad \left\{ \begin{array}{*{35}l} 2+i \\ i\text{ } \\ \end{array} \right.

We test the solutions:

\displaystyle \begin{align} & z=2+i:\quad z^{2}-2\left( 1+i \right)z+2i-1 \\ & =\left( 2+i \right)^{2}-2\left( 1+i \right)\left( 2+i \right)+2i-1 \\ & =4+4i+i^{2}-2\left( 2+i+2i+i^{2} \right)+2i-1 \\ & =4+4i-1-4-6i+2+2i-1=0 \\ & \\ \end{align}

\displaystyle \begin{align} & z=i:\quad z^{2}-2\left( 1+i \right)z+2i-1 \\ & =i^{2}-2\left( 1+i \right)i+2i-1 \\ & =-1-2\left( i+i^{2} \right)+2i-1 \\ & =-1-2i+2+2i-1=0 \\ \end{align}