If z_1=(2,-1) and z_2=(6,3),then operatorname | vedclass

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(A) Given $z_1 = 2 - i$ and $z_2 = 6 + 3i$. We need to find $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}\right)$. First,calculate $z_1 - z_2 = (2

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$-\frac{3 \pi}{4}-	an ^{-1}\left(\frac{1}{4}ight)$

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(A) Given $z_1 = 2 - i$ and $z_2 = 6 + 3i$. We need to find $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}ight)$. First,calculate $z_1 - z_2 = (2 - 6) + (-1 - 3)i = -4 - 4i$. Next,calculate $z_1 + z_2 = (2 + 6) + (-1 + 3)i = 8 + 2i$. Now,$\frac{z_1-z_2}{z_1+z_2} = \frac{-4-4i}{8+2i} = \frac{-2-2i}{4+i}$. Multiply numerator and denominator by the conjugate $(4-i)$: $\frac{(-2-2i)(4-i)}{(4+i)(4-i)} = \frac{-8 + 2i - 8i + 2i^2}{16 - i^2} = \frac{-8 - 6i - 2}{16 + 1} = \frac{-10 - 6i}{17} = -\frac{10}{17} - \frac{6}{17}i$. Alternatively,using the property $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}ight) = \operatorname{amp}(z_1-z_2) - \operatorname{amp}(z_1+z_2)$: $\operatorname{amp}(-4-4i) = -\pi + 	an^{-1}\left(\frac{-4}{-4}ight) = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$. $\operatorname{amp}(8+2i) = 	an^{-1}\left(\frac{2}{8}ight) = 	an^{-1}\left(\frac{1}{4}ight)$. Thus,the result is $-\frac{3\pi}{4} - 	an^{-1}\left(\frac{1}{4}ight)$.

If $z_1=(2,-1)$ and $z_2=(6,3)$,then $\operatorname{amp}\left(\frac{z_1-z_2}{z_1+z_2}\right)=$

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