If z = frac{(2-i)(1+i)^3}{(1-i)^2},then opera | vedclass

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

Question

(A) Given $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$.First,simplify $(1+i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i$.Next,simplify $(1-i)^2 = 1 - 2i

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{1}{3}ight) - \pi$

Vedclass · Answer

(A) Given $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$.First,simplify $(1+i)^3 = 1 + 3i + 3i^2 + i^3 = 1 + 3i - 3 - i = -2 + 2i$.Next,simplify $(1-i)^2 = 1 - 2i + i^2 = 1 - 2i - 1 = -2i$.Substituting these into $z$,we get $z = \frac{(2-i)(-2+2i)}{-2i} = \frac{(2-i) \cdot 2(-1+i)}{-2i} = \frac{(2-i)(1-i)}{i}$.Expanding the numerator: $(2-i)(1-i) = 2 - 2i - i + i^2 = 2 - 3i - 1 = 1 - 3i$.So,$z = \frac{1-3i}{i} = \frac{1-3i}{i} \cdot \frac{-i}{-i} = \frac{-i + 3i^2}{1} = -3 - i$.Since $z = -3 - i$ lies in the $3^{	ext{rd}}$ quadrant,the argument is $\operatorname{Arg}(z) = -\pi + 	an^{-1}\left(\frac{-1}{-3}ight) = -\pi + 	an^{-1}\left(\frac{1}{3}ight)$.

If $z = \frac{(2-i)(1+i)^3}{(1-i)^2}$,then $\operatorname{Arg}(z) = $

Similar Questions

Let $z$ and $w$ be complex numbers such that $\overline{z} + i\overline{w} = 0$ and $\text{arg}(zw) = \pi$. Then $\text{arg}(z)$ equals

The amplitude of $\sin \frac{\pi}{5} + i(1 - \cos \frac{\pi}{5})$ is:

Consider the following statements:
$I$: If $a$ and $b$ are positive real numbers,then $\sqrt{-a} \times \sqrt{-b} = \sqrt{ab}$
$II$: The argument of $\frac{1+i\sqrt{3}}{1-i\sqrt{3}}$ is $120^{\circ}$
Then:

$\operatorname{Arg}\left(\sin \frac{6 \pi}{5}+i\left(1+\cos \frac{6 \pi}{5}\right)\right)=$

If $\arg(z) < 0$,then $\arg(-z) - \arg(z)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module