If the complex number z = 2 - i(2 tan frac{5 | vedclass

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

Question

(A) Given $z = 2 - i(2 	an \frac{5 \pi}{8})$.Comparing with $z = x + iy$,we have $x = 2$ and $y = -2 	an \frac{5 \pi}{8}$.The modulus $r$ is given b

Vedclass · Accepted Answer

$(2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8})$

Vedclass · Answer

(A) Given $z = 2 - i(2 	an \frac{5 \pi}{8})$.Comparing with $z = x + iy$,we have $x = 2$ and $y = -2 	an \frac{5 \pi}{8}$.The modulus $r$ is given by $r = \sqrt{x^2 + y^2} = \sqrt{2^2 + (-2 	an \frac{5 \pi}{8})^2} = \sqrt{4(1 + 	an^2 \frac{5 \pi}{8})} = \sqrt{4 \sec^2 \frac{5 \pi}{8}} = |2 \sec \frac{5 \pi}{8}|$.Since $\frac{5 \pi}{8}$ is in the second quadrant,$\sec \frac{5 \pi}{8}$ is negative,so $r = -2 \sec \frac{5 \pi}{8} = 2 \sec(\pi - \frac{5 \pi}{8}) = 2 \sec \frac{3 \pi}{8}$.The argument $	heta$ is $	an^{-1}(\frac{y}{x}) = 	an^{-1}(\frac{-2 	an \frac{5 \pi}{8}}{2}) = 	an^{-1}(-	an \frac{5 \pi}{8}) = 	an^{-1}(	an(\pi - \frac{5 \pi}{8})) = 	an^{-1}(	an \frac{3 \pi}{8}) = \frac{3 \pi}{8}$.Thus,$(r, 	heta) = (2 \sec \frac{3 \pi}{8}, \frac{3 \pi}{8})$.

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

Similar Questions

If complex numbers $z_1$ and $z_2$ are such that $|z_1| = \sqrt{2}$,$|z_2| = \sqrt{3}$ and $|z_1 + z_2| = \sqrt{5 - 2\sqrt{3}}$,then the value of $|Arg(z_1) - Arg(z_2)|$ is

Convert the given complex number in polar form: $\sqrt{3}+i$

The amplitude of ${e^{{e^{ - i\theta }}}}$ is equal to

Let $z$ and $w$ be two complex numbers such that $\bar{z}+i \bar{w}=0$ and $\operatorname{Arg}(z w)=\pi$. Then,$\operatorname{Arg} z=$

If $z$ and $w$ are complex numbers such that $\bar{z} - i \bar{w} = 0$ and $\operatorname{Arg}(zw) = \frac{3 \pi}{4}$,then $\operatorname{Arg} z =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module