If for complex numbers z_1 and z_2,arg(z_1/z_ | vedclass

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(C) We know that $|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(	heta_1 - 	heta_2)$,where $	heta_1 = \arg(z_1)$ and $	heta_2 = \arg(z_2)$.G

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$||z_1| - |z_2||$

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(C) We know that $|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(	heta_1 - 	heta_2)$,where $	heta_1 = \arg(z_1)$ and $	heta_2 = \arg(z_2)$.Given that $\arg(z_1/z_2) = 0$,we have $\arg(z_1) - \arg(z_2) = 0$,which implies $	heta_1 - 	heta_2 = 0$.Substituting this into the equation:$|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2| \cos(0)$$|z_1 - z_2|^2 = |z_1|^2 + |z_2|^2 - 2|z_1||z_2|$$|z_1 - z_2|^2 = (|z_1| - |z_2|)^2$Taking the square root on both sides,we get $|z_1 - z_2| = ||z_1| - |z_2||$.

If for complex numbers $z_1$ and $z_2$,$\arg(z_1/z_2) = 0$,then $|z_1 - z_2|$ is equal to

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