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

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(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) =

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$\frac{-2}{5}$

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(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 $\operatorname{Re}\left(\frac{z_{1} z_{2}}{\bar{z}_{1}}\right)$.

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