Question

100 Баллов!!!! Решить 1 номер, 2 и 4.​

Answer 1

$1)\; \; (1+i)+(1-2i)=(1+1)+(i-2i)=2-i\\\\(6-5i)-(2-3i)=(6-2)+(-5i+3i)=4-2i\\\\(3+2i)(5-4i)=15-12i+10i-8i^2=15-2i+8=23-2i\\\\(1+i)^2=1+2i+i^2=1+2i-1=2i\\\\i(1+i)=i+i^2=i-1=-1+i$

$2)\; \; \frac{4+3i}{3+2i}=\frac{(4+3i)(3-2i)}{(3+2i)(3-2i)}=\frac{12-8i+9i-6i^2}{9-4i^2}=\frac{12+i+6}{9+4}=\frac{18+i}{13}=\frac{18}{13}+\frac{1}{13}\, i$

$4)\; \; a)\; \; z=3+4i\\\\|z|=\sqrt{3^2+4^2}=5\; ,\\\\\varphi =argz=arctg\frac{4}{3}\\\\z=5\cdot (cos\varphi +i\cdot sin\varphi )=5\cdot (\; cos(arctg\frac{4}{3})+i\cdot sin(arctg\frac{4}{3})\; )\\\\b)\; \; z=15-8i\\\\|z|=\sqrt{15^2+8^2}=\sqrt{289}=17\\\\\varphi =argz=arctg(-\frac{8}{15})\\\\z=17\cdot (cos(arctg(-\frac{8}{15})+i\cdot sin(arctg(-\frac{8}{15})\, )\\\\z=17\cdot (\, cos(-arctg\frac{8}{15})+i\cdot sin(-arctg\frac{8}{15})\, )\\\\c)\; \; z=2i\\\\|z|=2\; \; ,\; \; \varphi =argz=\frac{\pi}{2}$

$z=2\cdot (cos\frac{\pi}{2}+i\cdot sin\frac{\pi}{2})$

Answer 2

Ответ: во вложении Объяснение: