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}