Помогите решить высшая математика

z1=2+2i z2=3-4i z3=4+97i z4=1-22i z1 + z2 = 2+2i+3-4i = 5-2i z1-z2 = 2+2i -(3-4i) = -1+6i z2-z1 = 3-4i -(2+2i) = 1-6i [latex] \frac{z_1}{z_2}= \frac{2+2i}{3-4i}=\frac{(2+2i)(3+4i)}{(3-4i)(3+4i))}= \frac{6-8+6i+8i}{9+16}= \frac{-2+14i}{25}= -0,08 + 0,56i[/latex] [latex]\frac{z_2}{z_1}= \frac{3-4i}{2+2i}=\frac{(2-2i)(3-4i)}{(2+2i)(2-2i))}= \frac{6-8-6i-8i}{4+4}= \frac{-2-14i}{8}= -0,25 - 1,75i[/latex] z1+z2+z3+z4 = 2+2i+3-4i+4+97i+1-22i = 10+73i z4-z2 = 1-22i-(3-4i) = -2-18i z1*z2*z3*z4 =(2+2i)(3-4i)(4+97i)(1-22i) = (6+8+6i-8i)(4+2134-88i+97i)= =(14-2i)(2138+9i)=29932+18 -4276i+126i =29950 -4150i [latex] \frac{z_1+z_2}{z_4-z_3}= \frac{2+2i+3-4i}{1-22i-(4+97i)}= \frac{5-2i}{-3-119i}=\frac{(-5+2i)(3-119i)}{(3+119i)(3-119i)}= [/latex][latex]\frac{-15+238+6i+595i}{9+14161}=\frac{223+601i}{14170}= \frac{223}{14170}+ \frac{601}{14170}i [/latex]