Convert the given complex number in polar form $$\sqrt{3}+i$$

Asked by Abhisek | 1 year ago |  51

##### Solution :-

The complex number is $$\sqrt{3}+i$$

Let rcosθ = $$\sqrt{3}$$ and rsinθ = 1

$$r^2cos^2θ + r^2sin^2θ = (\sqrt{3})^2+1^2$$

$$r^2(cos^2θ + sin^2θ) =3+1$$

$$r^2=4$$

$$r=\sqrt{4}=2$$  [Conventionally, r > 0]

= 2cosθ = $$\sqrt{3}$$ and 2sinθ = 1

$$cosθ = \dfrac{\sqrt{3}}{2} and \;sinθ = \dfrac{1}{2}$$

$$θ =\dfrac{\pi}{6}$$  [As θ lies in the I quadrant]

$$\sqrt{3}+i= rcosθ + irsinθ$$

$$2 cos\dfrac{\pi}{6}+ i2sin\dfrac{\pi}{6}$$

$$=2(cos\dfrac{\pi}{6}+isin\dfrac{\pi}{6})$$

Answered by Pragya Singh | 1 year ago

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