(B) Let $z_1 = r(\cos \theta_1 + i \sin \theta_1)$.Given $|z_1| = |z_2|$,so $|z_2| = r$.Given $\text{amp}(z_1) + \text{amp}(z_2) = 0$,so $\text{amp}(z

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(B) Let $z_1 = r(\cos \theta_1 + i \sin \theta_1)$.Given $|z_1| = |z_2|$,so $|z_2| = r$.Given $\text{amp}(z_1) + \text{amp}(z_2) = 0$,so $\text{amp}(z_2) = -\theta_1$.Thus,$z_2 = r(\cos(-\theta_1) + i \sin(-\theta_1)) = r(\cos \theta_1 - i \sin \theta_1)$.Since $\bar{z}_1 = r(\cos \theta_1 - i \sin \theta_1)$,it follows that $\bar{z}_1 = z_2$.