If |z_1| = |z_2| and argleft( frac{z_1}{z_2} | vedclass

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(A) Given that $\arg\left( \frac{z_1}{z_2} \right) = \pi$. This implies $\arg(z_1) - \arg(z_2) = \pi$,or $\arg(z_1) = \arg(z_2) + \pi$. Let $\arg(z_2)

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(A) Given that $\arg\left( \frac{z_1}{z_2} ight) = \pi$. This implies $\arg(z_1) - \arg(z_2) = \pi$,or $\arg(z_1) = \arg(z_2) + \pi$. Let $\arg(z_2) = 	heta$. Then $\arg(z_1) = 	heta + \pi$. Since $|z_1| = |z_2|$,let $|z_1| = |z_2| = r$. We can write $z_2 = r(\cos 	heta + i \sin 	heta)$. Then $z_1 = r(\cos(	heta + \pi) + i \sin(	heta + \pi)) = r(-\cos 	heta - i \sin 	heta) = -r(\cos 	heta + i \sin 	heta) = -z_2$. Therefore,$z_1 + z_2 = -z_2 + z_2 = 0$.

If $|z_1| = |z_2|$ and $\arg\left( \frac{z_1}{z_2} \right) = \pi$,then $z_1 + z_2$ is equal to

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