The real part of the principal value of 2^{-i | vedclass

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

Question

(C) Let $z = 2^{-i}$.Taking the natural logarithm on both sides,we get:$\log z = \log (2^{-i})$$\Rightarrow \log z = -i \log 2$Since $z = e^{\log z}$,

Vedclass · Accepted Answer

$\cos \left[\log \left(\frac{1}{2}\right)\right]$

Vedclass · Answer

(C) Let $z = 2^{-i}$.Taking the natural logarithm on both sides,we get:$\log z = \log (2^{-i})$$\Rightarrow \log z = -i \log 2$Since $z = e^{\log z}$,we have:$z = e^{-i \log 2} = e^{i \log(1/2)}$Using Euler's formula,$e^{i 	heta} = \cos 	heta + i \sin 	heta$,where $	heta = \log(1/2)$:$z = \cos(\log(1/2)) + i \sin(\log(1/2))$The real part of $z$ is $\cos(\log(1/2))$.

The real part of the principal value of $2^{-i}$ is

Similar Questions

If $\theta = \frac{\pi}{12}$ and $x = \log \left(\cot \left(\frac{\pi}{4} + \theta\right)\right)$,then $\cosh x =$

Represent the complex number $z=1+i \sqrt{3}$ in the polar form.

$\sinh^{-1}\left(2^{3/2}\right)$ is equal to :

The real part of $(1 - i)^{-i}$ is

If the value of $\sqrt{-5-12 i}+\sqrt{7+24 i}$ is a negative real number $k$,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module