Find the modulus and the argument of the comp | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given complex number is $z = -\sqrt{3} + i$.Let $r \cos 	heta = -\sqrt{3}$ and $r \sin 	heta = 1$.On squaring and adding,we get:$r^{2}(\cos^{2}

Vedclass · Accepted Answer

Modulus $= 2$,Argument $= \frac{5\pi}{6}$

Vedclass · Answer

(A) Given complex number is $z = -\sqrt{3} + i$.Let $r \cos 	heta = -\sqrt{3}$ and $r \sin 	heta = 1$.On squaring and adding,we get:$r^{2}(\cos^{2} 	heta + \sin^{2} 	heta) = (-\sqrt{3})^{2} + (1)^{2}$$r^{2} = 3 + 1 = 4$$r = 2$ (since $r > 0$).Thus,the modulus is $2$.Now,$2 \cos 	heta = -\sqrt{3} \Rightarrow \cos 	heta = -\frac{\sqrt{3}}{2}$ and $2 \sin 	heta = 1 \Rightarrow \sin 	heta = \frac{1}{2}$.Since $\cos 	heta  0$,the angle $	heta$ lies in the $II$ quadrant.The reference angle is $\alpha = \frac{\pi}{6}$.Therefore,$	heta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.The modulus is $2$ and the argument is $\frac{5\pi}{6}$.

Find the modulus and the argument of the complex number $z = -\sqrt{3} + i$.

Similar Questions

Let $z$ satisfy $|z|=1$, $z=1-\bar{z}$ and $\operatorname{Im}(z) > 0$.
Statement-$I$: $z$ is a real number.
Statement-$II$: Principal argument of $z$ is $\frac{\pi}{3}$.
Then

Convert the following complex number into polar form: $\frac{1+7i}{(2-i)^2}$

The amplitude (argument) of $(1+i)^{5}$ is

Let $z = 1 + i$ and $z_1 = \frac{1 + i \overline{z}}{\overline{z}(1 - z) + \frac{1}{z}}$. Then $\frac{12}{\pi} \arg(z_1)$ is equal to $..........$.

Convert the given complex number in polar form: $1-i$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module