The argument of the complex number sin frac{6 | vedclass

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

Question

(C) Let $z = \sin \frac{6\pi}{5} + i(1 + \cos \frac{6\pi}{5})$.Using the identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and

Vedclass · Accepted Answer

$\frac{9\pi}{10}$

Vedclass · Answer

(C) Let $z = \sin \frac{6\pi}{5} + i(1 + \cos \frac{6\pi}{5})$.Using the identities $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ and $1 + \cos 	heta = 2 \cos^2 \frac{	heta}{2}$,we have:$z = 2 \sin \frac{3\pi}{5} \cos \frac{3\pi}{5} + i(2 \cos^2 \frac{3\pi}{5})$$z = 2 \cos \frac{3\pi}{5} (\sin \frac{3\pi}{5} + i \cos \frac{3\pi}{5})$Since $\cos \frac{3\pi}{5}$ is negative,we write $z = -2 \cos \frac{3\pi}{5} (-\sin \frac{3\pi}{5} - i \cos \frac{3\pi}{5})$.Using $-\sin 	heta = \cos(\frac{\pi}{2} + 	heta)$ and $-\cos 	heta = \sin(\frac{\pi}{2} + 	heta)$,we get:$z = |z| (\cos(\frac{\pi}{2} + \frac{3\pi}{5}) + i \sin(\frac{\pi}{2} + \frac{3\pi}{5}))$$z = |z| (\cos \frac{11\pi}{10} + i \sin \frac{11\pi}{10})$.Alternatively,the argument $	heta$ satisfies $	an 	heta = \frac{1 + \cos(6\pi/5)}{\sin(6\pi/5)} = \frac{2 \cos^2(3\pi/5)}{2 \sin(3\pi/5) \cos(3\pi/5)} = \cot(3\pi/5) = 	an(\frac{\pi}{2} - \frac{3\pi}{5}) = 	an(-\frac{\pi}{10})$.Since the real part is negative and the imaginary part is negative,the complex number lies in the $3^{rd}$ quadrant.Thus,$	heta = \pi - \frac{\pi}{10} = \frac{9\pi}{10}$.

The argument of the complex number $\sin \frac{6\pi}{5} + i(1 + \cos \frac{6\pi}{5})$ is

Similar Questions

The argument of $z = -1 - i\sqrt{3}$ is:

If $arg(z - a) = \frac{\pi}{4}$,where $a \in R$,then the locus of $z \in C$ is a

Let $z_1, z_2$ be two complex numbers such that $\bar{z}_1 - i \bar{z}_2 = 0$ and $\arg(z_1 z_2) = \frac{3 \pi}{4}$,then $\arg(z_1) =$

Convert the given complex number in polar form: $i$

If the complex number $z = 2 - i(2 \tan \frac{5 \pi}{8})$ has modulus $r$ and argument $\theta$,then what are $(r, \theta)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module