polar and trigonometric form?

Question

please answer me this question, i'm having trouble with it..i'm not good at math.

(4-2j) + (4-3j) + (1-2j)(1+2j) + -3+1j/2+1j + (5-3j) - (2+3j)

please give the answer as polar form and trigonometric form..
any1 hu could answer i thank you!

Answer 1

You used parentheses everywhere except on the quotient where they were needed. So I added them.

(4-2j) + (4-3j) + (1-2j)(1+2j) + (-3+j)/(2+j) + (5-3j) - (2+3j)

= 4 - 2j + 4 - 3j + (1 + 4) + (-1 + j) + 5 - 3j - 2 - 3j

= 15 - 10j

The magnitude r is:

r = √(15² + 10²) = √(225 + 100) = √325 = 5√13

The angle θ is

tanθ = -10/15 = -2/3

θ = arctan(-2/3)

In polar form the coordinates are

(r,θ) = [5√13, arctan(-2/3)]

Answer 2

=8-5j + (1-2j)(1+2j) + (-3+j)/(2+j) + (3-6j)
= (11 -11j) + (1-2j)(1-2j)/(1+2j)/(1-2j) + (-3+j)(2-j)/(2+j)/(2-j)
= (11-11j) + (1-4j+4j^2)/(1-4j^2) + (-6+5j-j^2)/(4-j^2)
= (11-11j) + (-3-4j)/(5) + (-5+5j)/(5)
= ...