Question

If $z_1=1+2i$ and $z_2=3+5i$, then the real part of $\frac{\overline{z_2}{z}_1}{z_2}$ is

• A. $-\frac{31}{17}$
• B. $\frac{17}{22}$
• C. $-\frac{17}{31}$
• D. $\frac{22}{17}$

Answer 1

The correct option is D

$\frac{22}{17}$

Explanation for the correct option.

Find the real part of $\frac{\overline{z_2}{z}_1}{z_2}$.

For the function $z_2=3+5i$ the conjugate is given as: $\overline{z_2}=3-5i$.

So substitute $z_1=1+2i$, $\overline{z_2}=3-5i$, and $z_2=3+5i$ in $\frac{\overline{z_2}{z}_1}{z_2}$ and simplify.

$\begin{array}{rcl}\frac{\overline{z_2}{z}_1}{z_2}& =& \frac{\left(3-5i\right)\left(1+2i\right)}{3+5i}\\ & =& \frac{3+6i-5i-10i^2}{3+5i}\\ & =& \frac{3+i-10\times \left(-1\right)}{3+5i}\left[i^2=-1\right]\\ & =& \frac{3+i+10}{3+5i}\\ & =& \frac{13+i}{3+5i}\end{array}$

Now multiply the numerator and denominator by $3-5i$.

$\begin{array}{rcl}\frac{\left(13+i\right)\left(3-5i\right)}{\left(3+5i\right)\left(3-5i\right)}& =& \frac{39-65i+3i-5i^2}{9-25i^2}\\ & =& \frac{39-62i-5\times \left(-1\right)}{9-25\times \left(-1\right)}\left[i^2=-1\right]\\ & =& \frac{39-62i+5}{9+25}\\ & =& \frac{44-62i}{34}\\ & =& \frac{22-31i}{17}\\ & =& \frac{22}{17}-\frac{31i}{17}\end{array}$

Thus, $\frac{\overline{z_2}{z}_1}{z_2}=\begin{array}{rcl}& & \frac{22}{17}\end{array}\begin{array}{rcl}& & -\end{array}\begin{array}{rcl}& & \frac{31i}{17}\end{array}$ and its real part is $\frac{22}{17}$.

Hence, the correct option is D.