Aus Online Mathematik Brückenkurs 2

Complete the square of the left-hand side:

\displaystyle \begin{align} & \left( z-\frac{2-i}{2} \right)^{2}-\left( \frac{2-i}{2} \right)^{2}+3-i=0 \\ & \left( z-\frac{2-i}{2} \right)^{2}-\left( 1-i+\frac{1}{4}i^{2} \right)+3-i=0 \\ & \left( z-\frac{2-i}{2} \right)^{2}-1+i+\frac{1}{4}+3-i=0 \\ & \left( z-\frac{2-i}{2} \right)^{2}+\frac{9}{4}=0 \\ & \\ \end{align}

Taking the root then gives that the solutions are

\displaystyle z-\frac{2-i}{2}=\pm \frac{3}{2}i\quad \Leftrightarrow \quad z=\left\{ \begin{array}{*{35}l} 1+i \\ 1-2i\text{ } \\ \end{array} \right.

Finally, we substitute the solutions into the equation and check that it is satisfied:

\displaystyle \begin{align} & z=1+i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\ & =\left( 1+i \right)^{2}-\left( 2-i \right)\left( 1+i \right)+3-i \\ & =1+2i+i^{2}-\left( 2+2i-i-i^{2} \right)+3-i \\ & =1+2i-1-2-i-1+3-i=0, \\ \end{align}

\displaystyle \begin{align} & z=1-2i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\ & =\left( 1-2i \right)^{2}-\left( 2-i \right)\left( 1-2i \right)+3-i \\ & =1-4i+4i^{2}-\left( 2-4i-i+2i^{2} \right)+3-i \\ & =1-4i-4-2+5i+2+3-i=0. \\ \end{align}