If z_1 = 5 - 2i and z_2 = 3 + i,where i = sqr | vedclass

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(A) Given $z_1 = 5 - 2i$ and $z_2 = 3 + i$. First,calculate $z_1 + z_2 = (5 + 3) + (-2 + 1)i = 8 - i$. Next,calculate $z_1 - z_2 = (5 - 3) + (-2 - 1)i

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$	an^{-1}\left(\frac{22}{19}ight)$

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(A) Given $z_1 = 5 - 2i$ and $z_2 = 3 + i$. First,calculate $z_1 + z_2 = (5 + 3) + (-2 + 1)i = 8 - i$. Next,calculate $z_1 - z_2 = (5 - 3) + (-2 - 1)i = 2 - 3i$. Now,consider the ratio $\frac{z_1 + z_2}{z_1 - z_2} = \frac{8 - i}{2 - 3i}$. To simplify,multiply the numerator and denominator by the conjugate of the denominator $(2 + 3i)$: $\frac{8 - i}{2 - 3i} 	imes \frac{2 + 3i}{2 + 3i} = \frac{16 + 24i - 2i - 3i^2}{2^2 + 3^2} = \frac{16 + 22i + 3}{4 + 9} = \frac{19 + 22i}{13} = \frac{19}{13} + \frac{22}{13}i$. The argument is given by $	an^{-1}\left(\frac{	ext{imaginary part}}{	ext{real part}}ight) = 	an^{-1}\left(\frac{22/13}{19/13}ight) = 	an^{-1}\left(\frac{22}{19}ight)$.

If $z_1 = 5 - 2i$ and $z_2 = 3 + i$,where $i = \sqrt{-1}$,then $\arg \left(\frac{z_1 + z_2}{z_1 - z_2}\right)$ is

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