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3 - 3i

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

z = |z|(cos θ + i sin θ)

let z = 3 - 3i in the form a + bi, where a = |z|cos θ and b = |z| sin θ

so that

|z| = √(a2 + b2) = √[(3)2 + (- 3)2] = √(9 + 9) = √18 = 3/√2

cos θ = a/|z| = 3/3√2 = 1/√2 and

sin θ = b/|z| = -3/3√2 = -1/√2

z = 3 - 3i = |z|(cos θ + i sin θ) =3√2(1/√2 - i 1/√2)

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14y ago

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