Question

Помогите решить упражнения над комплексными числами, заранее благодарен!

Answer 1

1.
a)
i² = -1
(1+i√3)/(1-i√3)  * (1+i√3)/(1+i√3)
= (1+i√3)²/(1² - i²*3 ) = (1+2i√3 + i²*3)/4 = (2i√3 -2)/4 = (i√3 -1)/2
б)
(i - 1)³/(i¹² + i¹³)
i¹² = (i²)⁶ = (-1)⁶ = 1
i¹³ = i¹²*i = 1*i = i
(i - 1)³/(i¹² + i¹³) = (i - 1)³/(1 + i) * (1 - i)/(1 - i)=(i - 1)³(1 - i)/2 = (i³ -3i² + 3i -1)(1 - i)/2 = = (-i + 3 + 3i -1)(1 - i)/2 = (2i + 2)(1 - i)/2 = (i + 1)(1 - i) = 1 - i²  = 1 -(-1) = 2
2.
a)
24(cos(75)  + i*sin(75))/3(cos(30)  + i*sin(30))
(cos(75)  + i*sin(75))/(cos(30) + i*sin(30)) * (cos(30) -i*sin(30))/(cos(30) -i*sin(30))= = (cos(75)  + i*sin(75))(cos(30)  - i*sin(30))/(cos²(30)  - i²*sin²(30)) =
= (cos(75)  + i*sin(75))(cos(30)  - i*sin(30))/(cos²(30)  + sin²(30)) =
= (cos(75)  + i*sin(75))(cos(30)  - i*sin(30)) =
= cos(75)cos(30)- i*sin(30)cos(75)+ i*sin(75)cos(30) - i²*sin(30)sin(75) =
= cos(75)cos(30)+ sin(30)sin(75) + i(sin(75)cos(30) - sin(30)cos(75)) =
= cos(75 - 30) + i*sin(75 - 30) = cos(45) + i*sin(45) = √2/2 + i*√2/2 = √2/2 *(1+ i)
 (24/3)* (√2/2) * (1 + i) = 6√2 * (1 + i)
б)
[e^(i*π/3)]/(-3 + i)⁵ =
= [e^(i*π/3)*(i + 3)⁵ ]/(i - 3)⁵(i + 3)⁵=[e^(i*π/3)*(i + 3)⁵ ]/(i² - 3²)⁵ =
= [e^(i*π/3) * (i + 3)⁵ ]/(i² - 3²)⁵  = [e^(i*π/3)*(i + 3)⁵ ]/(-10⁵ ) 
e^(i*π/3) = cos(π/3)  + i*sin(π/3) = 1/2 + i√3/2 = (1 + i√3)/2
(i + 3)⁴ = (i² + 6i + 9)² = (6i + 8)² = 4(3i + 4)² = 4(9i² + 24i + 16) = 4(7 +24i)
(i + 3)⁵ = (i + 3)⁴ (i + 3) = 4(7 +24i)(i + 3) = 4(7i +21 -24 + 72i) = 4(79i -3)
e^(i*π/3)*(i+3)^5 = 4(79i -3)*(1 + i√3)/2 = 2(79i -3)*(1 + i√3) = 2(79i -79√3 -3 -i3√3)=
= 2(-79√3 - 3 + i(79 - 3√3 ))
[e^(i*π/3)]/(-3 + i)⁵ = 2(-79√3 - 3 + i(79 - 3√3 ))/(-10⁵ ) =
= 2*(79√3 + 3 - i(79 - 3√3 ))/10⁵