Question

D ECE 341R Homework Assignmer X+ xythos.prod/584b1cebbb41a/3051253r t-disposition inline%3 ECE 341R-HomeworkAssignment#3 F_y 18, 201, 1. Perform the following operations and express the final result in the =x+ (also known as rectangular) form: (+30+1(-1+2)+(7-50). 2. Perform the following operations and express the final result in the 3. Find the polar (re) representation for the following complex numbers 4. For the problem in question number 2, express each term in parantheses in the (also known as rectangular) form: (4+2/12-3 (a) i2j +1 and (b) +3y. polar re form and then carry out the multiplication. (Your answer must match with the result you obtained for the answer to question number 2, above.) 5. In the continued product of complex numbers (2-DC-3+2D(5-40. express each of the complex numbers in parantheses in the exponential (or polar re'0 form) and carry out the multiplication. Vernify your result by direct straightforward multiplication Find the absolute value of the following complex numbers 0 (0+2)'; (6) 2+31 6. 1-j 7. Prove that ifx+y is a complex number, and is its complex conjugate, 8. Euler's thcorem (derived in class) states thatcos@jsin@. Employing this relationship, prove/derive the following well-known trigonometric identity a) sin20 + cos20 1 b) tan θ 1+i 0228 1+cos 26 Sote For both parts (a) and (b) use of Fuler's and/or Demoivre's theocem needs to be

Answer 1

(5 + 3j) + {(- 1 + 2j) + (7 - 5j)} = (5 + 3j) + (6 - 3j) = 11 + 0 j (4 + 2j) (2 - 3j) = 4(2 - 3j) + 2j (2 - 3j) = 8 - 12j + 4j - 6j^2 = 8 - 8j + 6 (Because j^2 = -1) = 14 - 8j a) j^2 + 2j + 1 = -1 + 2j + 1 = 2j as j^2 = -1 Here real part = x = 0 imaginary part = y = 2 = 0 + 2j Therefore r = squareroot x^2 + y^2 = squareroot 0 + 2^2 = squareroot 2^2 = 2 theta = tan^- 1 y/x = tan^- 1 2/0 = tan^- 1 infinity = pi/2 Therefore j^2 + 2j + 1 = re^i theta = 2e^pi/2 i b) (j + squareroot 3)^2 = j^2 + 2 squareroot 3 j + 3 = -1 + 2 squareroot 3 j + 3 = 2 + 2 squareroot 3 j Here x = 2 y = 2 squareroot 3 Therefore r = squareroot x^2 + y^2 = squareroot 4 + 12 = squareroot 16 = 4 theta = tan^- 1 y/x = tan^- 1 2 squareroot 3/3 = tan^- 1 squareroot 3 = pi/3 Therefore (j + squareroot 3)^2 = re^i theta = 4 e^(pi/3 i)