Question

(1+3i/1+i)^2 срочно нужно

Answer 1

$\Big (\frac{1+3i}{1+i}\Big )^2=\Big (\frac{(1+3i)(1-i)}{(1+i)(1-i)}\Big )^2=\Big (\frac{1-i+3i-3i^2}{1-i^2}\Big )^2=\Big (\frac{1+2i+3}{1+1}\Big )^2=\\\\=\Big (\frac{4+2i}{2}\Big )^2=(2+i)^2=2^2+2\cdot 2\cdot i+i^2=4+4i-1=3+4i$