Lösung 3.1:2d

Because the expression is rather large, we work step by step. We start by making the numerator and denominator each have the same denominator:

\displaystyle \begin{align}5-\frac{1}{1+i}&=\frac{5\cdot (1+i)}{1+i}-\frac{1}{1+i} = \frac{5+5i-1}{1+i} = \frac{4+5i}{1+i},\\ 3i+\frac{i}{2-3i} &= \frac{3i(2-3i)}{2-3i}+\frac{i}{2-3i}=\frac{6i-9i^2+i}{2-3i}=\frac{9+7i}{2-3i}.\end{align}

Hence,

\displaystyle \frac{5-\frac{1}{1+i}}{3i+\frac{i}{2-3i}}=\frac{\frac{4+5i}{1+i}}{\frac{9+7i}{2-3i}}=\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}.

We multiply out the numerator and denominator

\displaystyle \begin{align}\frac{(4+5i)(2-3i)}{(9+7i)(1+i)}&= \frac{4\cdot 2 -4 \cdot 3i +5i\cdot 2 - 5i\cdot 3i}{9\cdot1+9\cdot i +7i\cdot 1 +7i \cdot i}\\ &=\frac{8-12i+10i+15}{9+9i+7i-7}\\ &=\frac{23-2i}{2+16i}.\end{align}

This is an ordinary quotient of two complex numbers which we compute by multiplying top and bottom by the complex conjugate of the numerator:

\displaystyle \begin{align}\frac{23-2i}{2+16i}&=\frac{(23-2i)(2-16i)}{(2+16i)(2-16i)}\\ &=\frac{23\cdot 2 -23\cdot 16i -2i\cdot 2 +2i \cdot 16i}{2^2-(16i)^2}\\ &=\frac{46-368i-4i-32}{4+256}\\ &=\frac{14-372i}{260}\end{align}

If we divide up the numbers into factors,

\displaystyle \begin{align}14 &= 2\cdot 7\\ 372 &= 2\cdot 186=2\cdot 2\cdot 93=2\cdot 2\cdot 3\cdot 31\\ 260 &= 10\cdot 26=2\cdot 5\cdot 13\cdot 2\end{align}

we can simplify the answers:

\displaystyle \begin{align}\frac{14}{260}-\frac{372}{260}i&=\frac{2\cdot 7}{2\cdot 2\cdot 5\cdot 13}-\frac{2\cdot 2\cdot 3\cdot 31}{2\cdot 2\cdot 5\cdot 13}i\\ &=\frac{7}{2\cdot 5\cdot 13}-\frac{3\cdot 31}{5\cdot 13}i\\ &=\frac{7}{130}-\frac{93}{65}i\end{align}