Lösung 3.1:4b

If we divide both sides by 2-i, we obtain z by itself on the left-hand side:

\displaystyle z=\frac{3+2i}{2-i}.

It remains to calculate the quotient on the right-hand side. We multiply top and bottom by the complex conjugate of the numerator:

\displaystyle \begin{align}z&= \frac{(3+2i)(2+i)}{(2-i)(2+i)}=\frac{3\cdot 2+3\cdot i +2i\cdot 2+2i\cdot i}{2^2-i^2}\\ &=\frac{6+3i+4i-2}{4+1}=\frac{4+7i}{5}=\frac{4}{5}+\frac{7}{5}i\end{align}

Also, we substitute into the original equation to assure ourselves that we have calculated correctly:

\displaystyle \begin{align}LHS &= (2-i)z=(2-i)(\frac{4}{5}+\frac{7}{5}i)\\ &=2\cdot\frac{4}{5}-i\cdot\frac{4}{5}+2\cdot \frac{7}{5}i -i\cdot \frac{7}{5}i\\ &=\frac{8}{5}-\frac{4}{5}i+\frac{14}{5}i+\frac{7}{5} = \frac{8+7}{5}+\frac{14-4}{5}i\\ &=\frac{15}{5}+\frac{10}{5}i=3+2i=RHS.\end{align}