Aus Online Mathematik Brückenkurs 2

We complete the square on the left-hand side:

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

Taking the root now gives \displaystyle z+1=\pm i\sqrt{2} i.e. \displaystyle z=-1+i\sqrt{2} and \displaystyle z=-1-i\sqrt{2}.

We test the solutions in the equation to ascertain that we have calculated correctly.

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

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