If z_1=1-2 i,z_2=1+i,and z_3=3+4 i,then left( | vedclass

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(D) Given: $z_1=1-2 i, z_2=1+i, z_3=3+4 i$. We need to evaluate $\left(\frac{1}{z_1}+\frac{3}{z_2}\right) \frac{z_3}{z_2}$. First,calculate the term i

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$\frac{13}{2}-3 i$

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(D) Given: $z_1=1-2 i, z_2=1+i, z_3=3+4 i$. We need to evaluate $\left(\frac{1}{z_1}+\frac{3}{z_2}ight) \frac{z_3}{z_2}$. First,calculate the term inside the bracket: $\frac{1}{1-2 i}+\frac{3}{1+i} = \frac{(1+i)+3(1-2 i)}{(1-2 i)(1+i)} = \frac{1+i+3-6 i}{1+i-2 i-2 i^2} = \frac{4-5 i}{1-i+2} = \frac{4-5 i}{3-i}$. Now,multiply by $\frac{z_3}{z_2}$: $\left(\frac{4-5 i}{3-i}ight) \left(\frac{3+4 i}{1+i}ight) = \frac{(4-5 i)(3+4 i)}{(3-i)(1+i)} = \frac{12+16 i-15 i-20 i^2}{3+3 i-i-i^2} = \frac{12+i+20}{3+2 i+1} = \frac{32+i}{4+2 i}$. Rationalize the denominator: $\frac{32+i}{4+2 i} 	imes \frac{4-2 i}{4-2 i} = \frac{128-64 i+4 i-2 i^2}{16-4 i^2} = \frac{128-60 i+2}{16+4} = \frac{130-60 i}{20} = \frac{13}{2}-3 i$.

If $z_1=1-2 i$,$z_2=1+i$,and $z_3=3+4 i$,then $\left(\frac{1}{z_1}+\frac{3}{z_2}\right) \frac{z_3}{z_2}=$

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