Let Z = 4√2i - 4 = r(cosθ + isinθ)

Now, separating real and complex part , we get

-4 = rcosθ ……….eq.1

4√3 = rsinθ …………eq.2

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

64 = r²

Since r is always a positive no., therefore,

r = 8

hence its modulus is 8.

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

Since , and . therefore the lies in second the quadrant.

Tanθ = -√3, therefore θ= .

Representing the complex no. in its polar form will be

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