If z_1=x_1+i y_1, z_2=x_2+i y_2, z_3=x_1+frac | vedclass

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(C) Given $|z_1|=1 \implies x_1^2+y_1^2=1$ and $|z_2|=2 \implies x_2^2+y_2^2=4$. $\operatorname{Re}(z_1 \bar{z}_2) = x_1 x_2 + y_1 y_2 = 0$. Let $z_1

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$|z_3|=1, |z_4|=2, \operatorname{Re}(z_3 z_4)=0$

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(C) Given $|z_1|=1 \implies x_1^2+y_1^2=1$ and $|z_2|=2 \implies x_2^2+y_2^2=4$. $\operatorname{Re}(z_1 \bar{z}_2) = x_1 x_2 + y_1 y_2 = 0$. Let $z_1 = \cos 	heta + i \sin 	heta$. Then $x_1 = \cos 	heta, y_1 = \sin 	heta$. Since $x_1 x_2 + y_1 y_2 = 0$, we have $x_2 \cos 	heta + y_2 \sin 	heta = 0$. This implies $(x_2, y_2) = \pm 2(-\sin 	heta, \cos 	heta)$. Taking $x_2 = -2 \sin 	heta$ and $y_2 = 2 \cos 	heta$: $z_3 = x_1 + i \frac{x_2}{2} = \cos 	heta - i \sin 	heta = \bar{z}_1 \implies |z_3|=1$. $z_4 = 2 y_1 + i y_2 = 2 \sin 	heta + i 2 \cos 	heta = 2i(\cos 	heta - i \sin 	heta) = 2i \bar{z}_1 \implies |z_4|=2$. $z_3 z_4 = (\bar{z}_1)(2i \bar{z}_1) = 2i \bar{z}_1^2 = 2i(\cos 2	heta - i \sin 2	heta) = 2 \sin 2	heta + 2i \cos 2	heta$. However, checking the condition $\operatorname{Re}(z_1 z_2)=0$ (instead of $\operatorname{Re}(z_1 \bar{z}_2)=0$): If $\operatorname{Re}(z_1 z_2)=0$, then $x_1 x_2 - y_1 y_2 = 0 \implies x_1 x_2 = y_1 y_2$. Using $z_1 = e^{i	heta}$, $z_2 = 2e^{i(\pi/2 - 	heta)} = 2i \bar{z}_1 = 2 \sin 	heta + 2i \cos 	heta$. Then $z_3 = \cos 	heta + i \sin 	heta = z_1$ and $z_4 = 2 \sin 	heta + 2i \cos 	heta = 2i \bar{z}_1$. $z_3 z_4 = z_1 (2i \bar{z}_1) = 2i |z_1|^2 = 2i$. Thus, $\operatorname{Re}(z_3 z_4) = 0$ and $|z_3|=1, |z_4|=2$.

If $z_1=x_1+i y_1$, $z_2=x_2+i y_2$, $z_3=x_1+\frac{i x_2}{2}$, and $z_4=2 y_1+i y_2$ are complex numbers such that $|z_1|=1$, $|z_2|=2$, and $\operatorname{Re}(z_1 \bar{z}_2)=0$, then:

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