If $z_{1} = 1 - 2 i, z_{2} = 1 + i and z_{3} = 3 + 4 i$ , then $(\frac{1}{z_{1}} + \frac{3}{z_{2}}) \frac{z_{3}}{z_{2}} =$

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$6 - \frac{13}{2} i$

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

$13 - 3 i$

$13 - 6 i$

answer is D.

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Detailed Solution

Given $z_{1} = 1 - 2 i, z_{2} = 1 + i and z_{3} = 3 + 4 i$

Consider

$\begin{array}{rcl} (\frac{1}{z_{1}} + \frac{3}{z_{2}}) (\frac{z_{3}}{z_{2}}) & = & (\frac{1}{1 - 2 i} + \frac{3}{1 + i}) \frac{3 + 4 i}{1 + i} \\ = & (\frac{1 + 2 i}{5} + \frac{3 (1 - i)}{2}) (\frac{(3 + 4 i) (1 - i)}{2}) \\ = & (\frac{2 + 4 i + 15 - 15 i}{10}) (\frac{3 - 3 i + 4 i + 4}{2}) \\ = & (\frac{17 - 11 i}{10}) (\frac{7 + i}{2}) \\ = & \frac{119 + 17 i - 77 i + 11}{20} \\ = & \frac{130 - 60 i}{20} \\ = & \frac{13}{2} - 3 i \end{array}$