The argument and modulus of frac{1 + i}{1 - i | vedclass

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

Question

(D) Let $z = \frac{1 + i}{1 - i}$.Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$:$z = \frac{(1 + i)(1 + i)}{(

Vedclass · Accepted Answer

$\frac{\pi}{2}$ and $1$

Vedclass · Answer

(D) Let $z = \frac{1 + i}{1 - i}$.Multiplying the numerator and denominator by the conjugate of the denominator $(1 + i)$:$z = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1^2 - i^2} = \frac{1 + 2i - 1}{1 + 1} = \frac{2i}{2} = i$.We can write $z = 0 + 1i$.The modulus is $|z| = \sqrt{0^2 + 1^2} = 1$.The argument $	heta$ satisfies $\cos 	heta = 0$ and $\sin 	heta = 1$,which gives $	heta = \frac{\pi}{2}$.Thus,the argument is $\frac{\pi}{2}$ and the modulus is $1$.

The argument and modulus of $\frac{1 + i}{1 - i}$ are respectively:

Similar Questions

If $z = 1 + \cos \theta - i \sin \theta$ and $0 < \theta < \pi$,then $\left[|z - 1|^2 - \frac{|z|^2}{4}\right]^{1/2} =$

The number of non-zero integral solutions of the equation $|1 - i|^x = 2^x$ is

For any complex number $z$,$\bar{z} = \frac{1}{z}$ if and only if

If $m_1, m_2, m_3$ and $m_4$ respectively denote the moduli of the complex numbers $1+4i, 3+i, 1-i$ and $2-3i$,then the correct relation among the following is:

The Cartesian form of the complex number $z = 4(\cos 300^{\circ} + i \sin 300^{\circ})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module