If z_1 = 1+2i and z_2 = 3+5i,then text{Re} le | vedclass

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

Question

(D) Given,$z_1 = 1+2i$,$z_2 = 3+5i$,and $\bar{z}_2 = 3-5i$. First,calculate the numerator: $\bar{z}_2 z_1 = (3-5i)(1+2i) = 3 + 6i - 5i - 10i^2 = 3 + i

Vedclass · Accepted Answer

$\frac{22}{17}$

Vedclass · Answer

(D) Given,$z_1 = 1+2i$,$z_2 = 3+5i$,and $\bar{z}_2 = 3-5i$. First,calculate the numerator: $\bar{z}_2 z_1 = (3-5i)(1+2i) = 3 + 6i - 5i - 10i^2 = 3 + i + 10 = 13+i$. Now,calculate the expression: $\frac{\bar{z}_2 z_1}{z_2} = \frac{13+i}{3+5i}$. Multiply the numerator and denominator by the conjugate of the denominator $(3-5i)$: $\frac{13+i}{3+5i} 	imes \frac{3-5i}{3-5i} = \frac{39 - 65i + 3i - 5i^2}{3^2 + 5^2} = \frac{39 - 62i + 5}{9 + 25} = \frac{44 - 62i}{34}$. Simplifying the fraction: $\frac{44}{34} - \frac{62}{34}i = \frac{22}{17} - \frac{31}{17}i$. Therefore,$	ext{Re} \left( \frac{\bar{z}_2 z_1}{z_2} ight) = \frac{22}{17}$.

If $z_1 = 1+2i$ and $z_2 = 3+5i$,then $\text{Re} \left( \frac{\bar{z}_2 z_1}{z_2} \right) = $

Similar Questions

If $\frac{z - i}{z + i}$ $(z \ne -i)$ is a purely imaginary number,then $z \cdot \bar{z}$ is equal to

If $\alpha$ and $\beta$ are different complex numbers with $|\beta|=1$,then $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ is equal to

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

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

The product of two complex numbers,each of unit modulus,is also a complex number of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module