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 + i10 + i20 + i30 is a real number?
Solve the quadratic equations by factorization method only 6x2 – 17ix – 12 = 0
Solve the quadratic equations by factorization method only x2 + (1 – 2i)x – 2i = 0
Solve the quadratic equations by factorization method only x2 + 10ix – 21 = 0
Solve the quadratic equations by factorization method only 17x2 – 8x + 1 = 0