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

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(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}

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$\frac{22}{17}$

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(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

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