Argument of - 1 - isqrt 3 is | vedclass

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

Question

(d) Let $z = - 1 - i\sqrt 3 $
then $\alpha = {	an ^{ - 1}}\left| {\,\frac{b}{a}\,} ight| = {	an ^{ - 1}}\left| {\, - \frac{{\sqrt 3 }}{1}\,} ig

Vedclass · Accepted Answer

$ - \frac{{2\pi }}{3}$

Vedclass · Answer

(d) Let $z = - 1 - i\sqrt 3 $
then $\alpha = {	an ^{ - 1}}\left| {\,\frac{b}{a}\,} ight| = {	an ^{ - 1}}\left| {\, - \frac{{\sqrt 3 }}{1}\,} ight| = \frac{\pi }{3}$
Clearly, $z$ is in $III$ quadrant.
Therefore argument $	heta = - (\pi - \alpha ) = - (\pi - \pi /3) = \frac{{ - 2\pi }}{3}$.

Argument of $ - 1 - i\sqrt 3 $ is

Similar Questions

If $z = 3 + 5i,\,\,{\rm{then }}\,{z^3} + \bar z + 198 = $

If $\frac{{z - \alpha }}{{z + \alpha }}\left( {\alpha \in R} \right)$ is a purely imaginary number and $\left| z \right| = 2$, then a value of $\alpha $ is

The sum of amplitude of $z$ and another complex number is $\pi $. The other complex number can be written

Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0, z \in C$. Then $4\left(\alpha^2+\beta^2\right)$ is equal to :

If the conjugate of $(x + iy)(1 - 2i)$ be $1 + i$, then