Ребят помогите.Нужно сделать решение

1) i^15 = i^12*i^3 = (i^4)^3*i^3 = 1^3*(-i) = -i
cos 225 = cos(180 + 45) = -cos 45 = -√2/2
sin 225 = sin(180 + 45) = -sin 45 = -√2/2
$\frac{(5+3i)(5-3i)}{ \sqrt{2}(- \sqrt{2}/2-i* \sqrt{2}/2 ) } = \frac{25-9i^2}{-2/2-i*2/2}= \frac{25+9}{-1-i}=$
$= \frac{34(-1+i)}{(-1-i)(-1+i)}= \frac{34(-1+i)}{1-i^2} = \frac{34(-1+i)}{2} =-17+17i$

2) i^31 = i^28*i^3 = (i^4)^7*i^3 = 1^7*i^3 = 1(-i) = -i
i^0 = 1
(3 + 2i)^2 = 3^2 + 2*3*2i + 4i^2 = 9 + 12i - 4 = 5 + 12i
$\frac{(5+12i)(2-3*1)}{1-2i} = \frac{-(5+12i)}{1-2i} =- \frac{(5+12i)(1+2i)}{(1-2i)(1+2i)} =$
$= -\frac{5+12i+10i+24i^2}{1-4i^2}= -\frac{-19+22i}{5}= \frac{19}{5}- \frac{22}{5} i$