How do you divide (-4+2i)/(6-2i) in trigonometric form?

1 Answer
Sep 7, 2016

(sqrt(2)) / (2) (cos(0.412) - i sin(0.412))

Explanation:

We have: (- 4 + 2 i) / (6 - 2 i)

First, let's multiply both the numerator and the denominator by the complex conjugate of the denominator:

= (- 4 + 2 i) / (6 - 2 i) cdot (6 + 2 i) / (6 + 2 i)

= ((- 4) (6) + (- 4) (2 i) + (2 i) (6) + (2 i) (2 i)) / ((6)^(2) - (2 i)^(2))

= (- 24 - 8 i + 12 i + 4 i^(2)) / (36 - 4 i^(2))

Let's apply the fact that i^(2) = - 1:

= (- 24 + (4 cdot (- 1)) + 4 i) / (36 - (4 cdot (- 1)))

= (- 24 - 4 + 4 i) / (36 + 4)

= (- 28 + 4 i) / (40)

= - (7) / (10) + (1) / (10) i

Then, we need to determine the modulus and the argument of this complex number.

Let z = - (7) / (10) + (1) / (10) i:

=> |z| = sqrt((- (7) / (10))^(2) + ((1) / (10))^(2))

=> |z| = sqrt((49) / (100) + (1) / (100))

=> |z| = sqrt((50) / (100))

=> |z| = sqrt((1) / (2))

=> |z| = (sqrt(2)) / (2)

and

=> arg(z) = arctan(((1) / (10)) / (- (7) / (10)))

=> arg(z) = arctan(- (1) / (7))

=> arg(z) approx - 0.142

Now, let's express the complex number in polar form:

=> z = (sqrt(2)) / (2) (cos(- 0.412) + i sin(- 0.412))

=> z = (sqrt(2)) / (2) (cos(0.412) - i sin(0.412))