Let Z = 3√2i - 3√2 = r(cos + isinθ)

Now, separating real and complex part , we get

-3√2 = rcosθ ……….eq.1

3√2 = rsinθ …………eq.2

Squaring and adding eq.1 and eq.2, we get

36 = r²

Since r is always a positive no., therefore,

r = 6

hence its modulus is 6.

now, dividing eq.2 by eq.1, we get,

Since , and tanθ = -1 . therefore the θ lies in secothe nd quadrant.

Tanθ = -1 , therefore θ = .

Representing the complex no. in its polar form will be

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