The complex number is

\( z = -\sqrt{3} + i\)

Let rcosθ = \( -\sqrt{3}\) and rsinθ = 1

squaring and adding

\( (rcosθ)^2 + (rsinθ)^2 =(-\sqrt{3})^2+(-1)^2\)

\( r^2=3+1=4LLL\) \( [cos^2\theta+sin^2\theta=1]\)

\( r=\sqrt{4}\) = 2LLL[ Conventionally, r > 0]

Modulus = 2

\( 2cosθ=-\sqrt{3}\) and 2sinθ = 1

\(cosθ =\dfrac{-\sqrt{3}}{2}\) and sinθ = \( \dfrac{1}{2}\)

\( \theta=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}\)LL[As θ lies in the II quadrant]

Thus, the modulus and argument of the complex number \( -\sqrt{3} + i\) are 2 and \( \dfrac{5\pi}{6}\)

Answered by Abhisek | 1 year agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

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