The complex number is
\(z= – 1 – i\sqrt{3}\)
Let rcosθ = -1 and rsinθ = \( -\sqrt{3}\)
Squaring and adding
(rcosθ)2 + (rsinθ)2 = \((-1)^2+(- \sqrt{3})^2\)
\( r^2(cos^2θ) + (sin^2θ)=1+3\)
r2 = 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 3rd 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 + i10 + i20 + i30 is a real number?
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