If Z_1=2+i and Z_2=3-4i,and frac{overline{Z_1 | vedclass

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(B) Given $Z_1=2+i$ and $Z_2=3-4i$.The conjugates are $\overline{Z_1}=2-i$ and $\overline{Z_2}=3+4i$.We need to evaluate $\frac{\overline{Z_1}}{\overl

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$-1$

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(B) Given $Z_1=2+i$ and $Z_2=3-4i$.The conjugates are $\overline{Z_1}=2-i$ and $\overline{Z_2}=3+4i$.We need to evaluate $\frac{\overline{Z_1}}{\overline{Z_2}} = \frac{2-i}{3+4i}$.Multiply the numerator and denominator by the conjugate of the denominator $(3-4i)$:$\frac{2-i}{3+4i} 	imes \frac{3-4i}{3-4i} = \frac{6-8i-3i+4i^2}{3^2-(4i)^2}$.Since $i^2=-1$,we have $\frac{6-11i-4}{9+16} = \frac{2-11i}{25} = \frac{2}{25} - \frac{11}{25}i$.Comparing this with $a+bi$,we get $a=\frac{2}{25}$ and $b=\frac{-11}{25}$.Now,calculate $-7a+b = -7(\frac{2}{25}) - \frac{11}{25} = \frac{-14-11}{25} = \frac{-25}{25} = -1$.

If $Z_1=2+i$ and $Z_2=3-4i$,and $\frac{\overline{Z_1}}{\overline{Z_2}}=a+bi$,then the value of $-7a+b$ is (where $i=\sqrt{-1}$ and $a, b \in \mathbb{R}$)

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