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(5√3 - 5i)/(4 + 4i√3)

Let's try to represent the given complex numbers in the polar form.

z = |z|(cos θ + i sin θ)

Let z1 = 5√3 - 5i in the form a + bi, where

a1 = |z1|cos θ1 and

b1 = |z1| sin θ1

so that

|z1| = √(a12 + b12) = √[(5√3)2 + (- 5)2] = √(75 + 25) = √100 = 10

cos θ = a1/|z1|= 5√3/10 = √3/2 and

sin θ = b1/|z1|= -5/10 = -1/2

So that,

z1=5√3 - 5=|z1|(cos θ1 + i sin θ1) =10(√3/2 - i1/2), where θ1 = 330 degrees

Let z2 = 4 + 4√3i

|z2|=√(42 + (4√3)2) = √(16 + 48) = √64= 8

cos θ2 = 4/8 = 1/2

sin θ2 = (4√3)/8 = √3/2

So that,

z2 = 4 + 4√3 = |z2|(cos θ2 + i sin θ2) = 8(1/2 + √3/2i), where θ2 = 60 degrees

Now let's divide.

Recall the Euler formula: z = |z|eiθ, wher eiθ = (cos θ + i sin θ)

z1/z2 = |z1|eiθ1/|z2|eiθ2 = (|z|/|z|)ei(θ1 - θ2) = (|z|/|z|)[cos (θ1 - θ2) + i sin (θ1 - θ2)]

z1/z2 = (10/8)[cos (330 - 60) + i sin (330 - 60)] = (5/2)( cos 270 + i sin 270) = (5/4)(0 - i )

Let's check:

(5√3 - 5i)/(4 + 4i√3)

= (5√3 - 5i)(4 - 4i√3)/(4 + 4i√3)(4 - 4i√3)

= (20√3 - 60i - 20i + 20i2√3)/(16 - 48i2)

= (20√3 - 80i - 20√3)/(16 + 48)

= -80i/64

= -5i/4

= 0 - 5i/4

= (5/4)(0 - i)

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