For a, b, c, d in R, if z_1 = a + ib and z_2 | vedclass

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(A) Given $|z_1| = |z_2| = 1$, we can write $z_1 = \cos \alpha + i \sin \alpha$ and $z_2 = \cos \beta + i \sin \beta$.$\operatorname{Re}(z_1 \bar{z}_2

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$\operatorname{Re}(w_1 \bar{w}_2) = 0$

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(A) Given $|z_1| = |z_2| = 1$, we can write $z_1 = \cos \alpha + i \sin \alpha$ and $z_2 = \cos \beta + i \sin \beta$.$\operatorname{Re}(z_1 \bar{z}_2) = ac + bd = \cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos(\alpha - \beta) = 0$.This implies $\alpha - \beta = \pm \frac{\pi}{2}$.Now, for $w_1 = a + ic = \cos \alpha + i \cos \beta$ and $w_2 = b + id = \sin \alpha + i \sin \beta$, we calculate $\operatorname{Re}(w_1 \bar{w}_2) = ab + cd$.$\operatorname{Re}(w_1 \bar{w}_2) = \cos \alpha \sin \alpha + \cos \beta \sin \beta = \frac{1}{2}(\sin 2\alpha + \sin 2\beta)$.Using the sum-to-product formula, $\frac{1}{2}(2 \sin(\alpha + \beta) \cos(\alpha - \beta))$.Since $\cos(\alpha - \beta) = \cos(\pm \frac{\pi}{2}) = 0$, the expression equals $0$.Thus, $\operatorname{Re}(w_1 \bar{w}_2) = 0$.

For $a, b, c, d \in R$, if $z_1 = a + ib$ and $z_2 = c + id$ are such that $|z_1| = |z_2| = 1$ and $\operatorname{Re}(z_1 \bar{z}_2) = 0$, then the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfy

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