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Write 2i in polar form.

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Let, z = 2i

Let 0 = r cosθ and 2 = r sinθ

By squaring and adding, we get

(0)2 + (2)2 = (r cosθ)2 + (r sinθ)2

⇒ 0+4 = r2 (cos2θ + sin2θ)

⇒ 4 = r2

⇒ r = 2

∴ cosθ = 0 and sinθ = 1

Since, θ lies in first quadrant, we have

θ = π/2

Thus, the required polar form is \(2[cos(\frac{\pi}{2})+i\,sin(\frac{\pi}{2})]\)

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