Lösung 3.3:4b
Aus Online Mathematik Brückenkurs 2
Typically, one solves a second-degree by completing the square, followed by taking the root.
If we complete the square of the left-hand side, we get
\displaystyle \begin{align}
& \left( z-2 \right)^{2}-2^{2}+5=0, \\
& \left( z-2 \right)^{2}+1=0. \\
\end{align}
Taking the root then gives that the equation has roots
\displaystyle z-2=\pm i,
i.e.
\displaystyle z=\text{2}+i
and
\displaystyle z=\text{2}-i.
If we want to be sure that we have found the correct solutions, we can substitute each solution into the equation and see whether the equation is satisfied.
\displaystyle \begin{align}
& z=\text{2}+i:\quad z^{2}-4z+5=\left( \text{2}+i \right)^{2}-4\left( \text{2}+i \right)+5 \\
& =2^{2}+4i+i^{2}-8-4i+5 \\
& =4+4i-1-8-4i+5=0 \\
\end{align}
\displaystyle \begin{align}
& z=\text{2-}i:\quad z^{2}-4z+5=\left( \text{2-}i \right)^{2}-4\left( \text{2-}i \right)+5 \\
& =2^{2}-4i+i^{2}-8+4i+5 \\
& =4-4i-1-8+4i+5=0 \\
\end{align}