Question

Очень нужна от 4 задания до 5
( можно все ,но в приоритете эти за решение!​

Answer 1

$1)\; \; z_1=2-3i\; ,\; z_2=3+2i\; ,\; z_3=3-2i\\\\z_1+z_2=5-i\; \; ;\; \; z_1+z_3=5-5i\; \; ;\; \; z_1-z_2=-1-5i\; \; ;\; \; z_2-z_3=4i\; ;\\\\z_1\cdot z_2=(2-3i)(3+2i)=6-5i-6i^2=12-5i\\\\z_2\cdot z_3=(3+2i)(3-2i)=9-4i^2=9+4=13$

$2)\; \; (2-i)(2+i)-(3-2i)-7=4-i^2-3+2i-7=-5+2i\\\\\\3)\; \; \dfrac{1}{i}=\dfrac{i}{i^2}=-i\\\\\\\dfrac{1}{2+i}=\dfrac{2-i}{(2+i)(2-i)}=\dfrac{2-i}{4-i^2}=\dfrac{2-i}{5}=0,4-0,2i\\\\\\\dfrac{3-i}{2+2i}=\dfrac{(3-i)(2-2i)}{(2+2i)(2-2i)}=\dfrac{6-8i+2i^2}{4-4i^2}=\frac{6-8i}{4+4}=0,75-i$

$4)\; \; a)\; \; -3=3\, (cos\pi +i\, sin\pi )\\\\b)\; \; \; -i=cos(-\frac{\pi}{2})+i\, sin(-\frac{\pi}{2})\\\\c)\; \; \; z=1+i\; \; ,\; \; |z|=\sqrt{1^2+1^2}=\sqrt2\; ,\\\\$

$cos\varphi =\frac{1}{\sqrt2}\; \; ,\; \; sin\varphi =\frac{1}{\sqrt2}\; \; \to \; \; \; tg\varphi =\frac{1/\sqrt2}{1/\sqrt2}=1\\\\argz=\varphi =arctg1=\frac{\pi}{4}\; \; ,\; \; \varphi =\frac{\pi}{4}\\\\z=\sqrt2\, (cos\frac{\pi}{4}+i\, sin\frac{\pi}{4})\\\\d)\; \; z=-2+2i\; \; ,\; \; |z|=\sqrt{2^2+2^2}=2\sqrt2\\\\cos\varphi =\frac{-2}{2\sqrt2}=-\frac{\sqrt2}{2}\; \; ,\; \; sin\varphi =\frac{2}{2\sqrt2}=\frac{\sqrt2}{2}\; ,$

$tg\varphi =-1\; \; \Rightarrow \; \; \; \varphi =\frac{3\pi}{4}\\\\z=2\sqrt2\, (cos\frac{3\pi}{4}+i\, sin\frac{3\pi}{4})$