If i=sqrt{-1},then operatorname{Arg}left[frac | vedclass

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

Question

(A) Let $Z = \frac{(1+i)^{2025}}{(1-i)^{2022}}$.We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.Substituting these values:$Z

Vedclass · Accepted Answer

$\frac{-\pi}{4}$

Vedclass · Answer

(A) Let $Z = \frac{(1+i)^{2025}}{(1-i)^{2022}}$.We know that $1+i = \sqrt{2} e^{i\pi/4}$ and $1-i = \sqrt{2} e^{-i\pi/4}$.Substituting these values:$Z = \frac{(\sqrt{2} e^{i\pi/4})^{2025}}{(\sqrt{2} e^{-i\pi/4})^{2022}}$$Z = \frac{(\sqrt{2})^{2025} e^{i(2025\pi/4)}}{(\sqrt{2})^{2022} e^{-i(2022\pi/4)}}$$Z = (\sqrt{2})^3 e^{i(2025\pi/4 + 2022\pi/4)}$$Z = 2\sqrt{2} e^{i(4047\pi/4)}$Since $4047\pi/4 = 1011\pi + 3\pi/4$,the principal argument is $\operatorname{Arg}(Z) = \frac{3\pi}{4} - \pi = -\frac{\pi}{4}$.

If $i=\sqrt{-1}$,then $\operatorname{Arg}\left[\frac{(1+i)^{2025}}{(1-i)^{2022}}\right]=$

Similar Questions

If $z$ is a complex number such that $|z| = 4$ and $\text{arg}(z) = \frac{5\pi}{6}$,then $z$ is equal to

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

If $|z| = 4$ and $\text{arg}(z) = \frac{5\pi}{6}$,then $z =$

$\operatorname{Arg}\left(\frac{4+2 i}{1-2 i}+\frac{3+4 i}{2+3 i}\right)$ lies in the interval

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module