The complex number is

\(z= – 1 – i\sqrt{3}\)

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

Squaring and adding

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

\( r^2(cos^2θ) + (sin^2θ)=1+3\)

r^{2} = 4 \([ cos^2θ + sin^2θ=1]\)

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

Modulus = 2

2cosθ = -1 and 2sinθ = \( -\sqrt{3}\)

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

Since both the values of sinθ and cosθ negative and sinθ and cosθ are negative in 3^{rd} quadrant,

Argument = \( -(\pi-\dfrac{\pi}{3})=\dfrac{-2\pi}{3}\)

Thus, the modulus and argument of the complex number \( – 1 – i\sqrt{3}\) are 2 and \( \dfrac{-2\pi}{3}\) Respectively

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

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