Lösung 3.1:2c

We start by expanding the quadratic in the numerator with the square rule:

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

Then, we multiply top and bottom by the complex conjugate of the numerator:

\displaystyle \begin{align}\frac{1-4\sqrt{3}i}{1+i\sqrt{3}}&=\frac{(1-4\sqrt{3}i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}\\ &=\frac{1\cdot 1-1\cdot i\sqrt{3}-1\cdot 4\sqrt{3}i+ 4\sqrt{3}i\cdot i \sqrt{3}}{1^2-(i\sqrt{3})^2}\\ &=\frac{1-i\sqrt{3}-4\sqrt{3}i+4(\sqrt{3})^2i^2}{1+(\sqrt{3})^2)}\\ &=\frac{1-(\sqrt{3}+4\sqrt{3})i-4\cdot 3}{1+3}\\ &=\frac{1-12-(1+4)\sqrt{3}i}{4}\\ &=-\frac{11}{4}-\frac{5\sqrt{3}}{4}i.\end{align}