{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hwsolB01

# hwsolB01 - Therefore v = 1(cos(pi j*sin(pi =-1 j*0(v...

This preview shows pages 1–2. Sign up to view the full content.

Sheet1 Page 1 HW 1 SOLUTION _____________ (i) Plot (-1,sqrt(3)) and (1,-1) on the (x,y) plane. (ii) |z1| = sqrt((-1)^2+(sqrt(3))^2) = sqrt(1+3) = 2 angle(z1) = arctan(-sqrt(3)) + pi = 2*pi/3 = 2.0944 |z2| = sqrt(1+1) = sqrt(2) = 2.828 angle(z2) = arctan(-1/1) = -pi/4 (iii) z1^3 = (-1+sqrt(3)*j)*(-1+sqrt(3)*j)*(-1+sqrt(3)*j) = (-1+sqrt(3)*j)*(1-sqrt(3)*j-sqrt(3)*j+3*j^2) = (-1+sqrt(3)*j)*(-2-2*sqrt(3)*j) = 2-2*sqrt(3)*j+2*sqrt(3)*j-6*j^2 = 8 z2^2 = (1-j)*(1-j) = 1-j-j+j^2 = -2j z1^2 = (-1+sqrt(3)*j)*(-1+sqrt(3)*j) = 1-sqrt(3)*j-sqrt(3)*j+3*j^2 = -2-2*sqrt(3)*j z1^2 + 2*z1 + 4 = -2-2*sqrt(3)*j -2 +2*sqrt(3)*j + 4 = 0 (iv) v = (z1)^6*(z2)^(-12) |v| = |z1|^6*|z2|^(-12) = 2^6 * (sqrt(2))^(-12) = 1 angle(v) = 6*angle(z1) - 12*angle(z2) = 6*(2*pi/3) - 12*(-pi/4) = 4*pi+3*pi = 7*pi (same as pi = 3.14159265)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Therefore v = 1*(cos(pi) + j*sin(pi)) = -1 + j*0 (v) |z-z1-z2|=|z1| is a circle of radius |z1| = sqrt(1+3) = 2 with center at the point z1+z2 = (0,sqrt(3)-1). Note that z=z2 lies on that circle. If z is the point of intersection of the circle and the real axis, then z = x + j*0 = x (i.e., z lies on the real axis) Sheet1 Page 2 and |x + (sqrt(3)-1) *j| = 2 Squaring both sides, x^2 + (sqrt(3)-1)^2 = 4 which has roots x = -1.8612 and x = 1.8612 x If z is the point of intersection of the circle and the imaginary axis, then z = j*y (i.e., z lies on the imaginary axis) and |y*j + (sqrt(3)-1) *j| = 2 Squaring both sides, (y + sqrt(3)-1)^2 = 4 which has roots y = 2+1-sqrt(3) = 3-sqrt(3) and y = 2-1+sqrt(3) = 1+sqrt(3)...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

hwsolB01 - Therefore v = 1(cos(pi j*sin(pi =-1 j*0(v...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online