Aus Online Mathematik Brückenkurs 2

Before we can complete the square of the expression, we need to take out the factor \displaystyle i in front of \displaystyle z^{2}

\displaystyle i\left( z^{2}+\frac{2+3i}{i}z-\frac{1}{i} \right).

Then, simplify the complex fractions by multiplying top and bottom by \displaystyle -i (the denominator's complex conjugate):

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

Now we are ready to complete the square of the second-degree expression inside the bracket:

\displaystyle \begin{align} & i\left( z^{2}+\left( 3-2i \right)z+i \right)=i\left( \left( z+\frac{3-2i}{2} \right)^{2}-\left( \frac{3-2i}{2} \right)^{2}+i \right) \\ & =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\left( \frac{3}{2}-i \right)^{2}+i \right) \\ & =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\frac{9}{4}+3i+1+i \right) \\ & =i\left( \left( z+\frac{3}{2}-i \right)^{2}-\frac{5}{4}+4i \right) \\ & =\left( z+\frac{3}{2}-i \right)^{2}-\frac{5}{4}i+4i^{2} \\ & =\left( z+\frac{3}{2}-i \right)^{2}-4-\frac{5}{4}i. \\ \end{align}