Let z_{1} = 2 - i and z_{2} = -2 + i. Find op | vedclass

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

Question

(A) Given $z_{1} = 2 - i$ and $z_{2} = -2 + i$. First, calculate the product $z_{1} z_{2} = (2 - i)(-2 + i) = -4 + 2i + 2i - i^{2} = -4 + 4i - (-1) =

Vedclass · Accepted Answer

$\frac{-2}{5}$

Vedclass · Answer

(A) Given $z_{1} = 2 - i$ and $z_{2} = -2 + i$. First, calculate the product $z_{1} z_{2} = (2 - i)(-2 + i) = -4 + 2i + 2i - i^{2} = -4 + 4i - (-1) = -3 + 4i$. The conjugate $\bar{z}_{1} = 2 + i$. Now, consider the expression $\frac{z_{1} z_{2}}{\bar{z}_{1}} = \frac{-3 + 4i}{2 + i}$. To simplify, multiply the numerator and denominator by the conjugate of the denominator, $(2 - i)$: $\frac{(-3 + 4i)(2 - i)}{(2 + i)(2 - i)} = \frac{-6 + 3i + 8i - 4i^{2}}{2^{2} + 1^{2}} = \frac{-6 + 11i - 4(-1)}{4 + 1} = \frac{-6 + 11i + 4}{5} = \frac{-2 + 11i}{5} = \frac{-2}{5} + \frac{11}{5}i$. Comparing the real part, we get $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right) = \frac{-2}{5}$.

Let $z_{1} = 2 - i$ and $z_{2} = -2 + i$. Find $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)$.

Similar Questions

$\frac{1 - i}{1 + i}$ is equal to

The modulus of $\frac{1-i}{3+i}+\frac{4i}{5}$ is

$\left|\frac{1}{i^{2020}}+\frac{2}{i^{2021}}+\frac{3}{i^{2022}}+\frac{4}{i^{2023}}\right|$ is equal to

If complex numbers $(x - 2y) + i(3x - y)$ and $(2x - y) + i(x - y + 6)$ are conjugates of each other,then $|x + iy|$ is $(x, y \in \mathbb{R})$.

If $z$ is a complex number such that $\left|z-\frac{4}{z}\right|=2$,then the greatest value of $|z|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module