= i⁷⁵

= i⁴ⁿ⁺³ where n = 18

since i⁴ⁿ⁺³ = -i

i⁷⁵ = -i

Let Z = -i = r(cosθ + isinθ)

Now , separating real and complex part , we get

0 = rcosθ ……….eq.1

-1 = rsinθ …………eq.2

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

1 = r²

Since r is always a positive no., therefore,

r = 1,

hence its modulus is 1.

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

tanθ = -∞

Since cosθ = 0 , sinθ = -1 and tanθ = -∞ . therefore the θ lies in fourth quadrant.

Tanθ = -∞ , therefore

Representing the complex no. in its polar form will be

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