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

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

Question

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

Vedclass · Accepted Answer

$\frac{22}{17}$

Vedclass · Answer

(D) Given ${z_1} = 1 + 2i$ and ${z_2} = 3 + 5i$. First, find the conjugate of ${z_2}$: ${\bar{z}_2} = 3 - 5i$. Now, calculate the numerator: ${\bar{z}_2}{z_1} = (3 - 5i)(1 + 2i) = 3 + 6i - 5i - 10i^2 = 3 + i + 10 = 13 + i$. Next, divide by ${z_2}$: $\frac{13 + i}{3 + 5i}$. To simplify, multiply the numerator and denominator by the conjugate of the denominator $(3 - 5i)$: $\frac{13 + i}{3 + 5i} 	imes \frac{3 - 5i}{3 - 5i} = \frac{39 - 65i + 3i - 5i^2}{3^2 + 5^2} = \frac{39 - 62i + 5}{9 + 25} = \frac{44 - 62i}{34}$. Simplify the fraction: $\frac{44}{34} - \frac{62}{34}i = \frac{22}{17} - \frac{31}{17}i$. The real part $\operatorname{Re} \left( \frac{{\bar{z}_2}{z_1}}{{z_2}} ight)$ is $\frac{22}{17}$.

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

Similar Questions

If ${z_1}$ and ${z_2}$ are two complex numbers,then $Re({z_1}{z_2}) = $

The value of $(1+i)^5(1-i)^7$ is:

If $x+\frac{1}{x}=2 \sin \alpha$ and $y+\frac{1}{y}=2 \cos \beta$,then $x^3 y^3+\frac{1}{x^3 y^3}=$

The expression $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ equals

Let $\bar{z}$ denote the complex conjugate of a complex number $z$ and let $i=\sqrt{-1}$. In the set of complex numbers,the number of distinct roots of the equation $\bar{z}-z^2=i(\bar{z}+z^2)$ is . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module