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

Question

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

Delilah Aguilar · Accepted Answer

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

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

Let's start by multiplying the denominator and the numerator by the denominator's complex conjugate:

#= (- 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#

Next, we must ascertain this complex number's argument and modulus.

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#

It is now time to put the complex number into polar form:

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

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