A) \[-31/17\]
B) \[17/22\]
C) \[-17/31\]
D) \[22/17\]
Correct Answer: D
Solution :
We have, \[{{z}_{1}}=1+2i\] and \[{{z}_{2}}=3+5i\] \[\therefore \] \[{{\bar{z}}_{2}}=3-5i\] Now, \[{{\bar{z}}_{2}}{{z}_{1}}=(3-5i)(1+2i)\] \[=3+6i-5i-10{{i}^{2}}\] \[=3+i+10=13+i\] \[\therefore \] \[\frac{{{{\bar{z}}}_{2}}{{z}_{1}}}{{{z}_{2}}}=\frac{(13+i)}{(3+5i)}\times \frac{(3-5i)}{(3-5i)}\] \[=\frac{39-65i+3i-5{{i}^{2}}}{9+25}\] \[=\frac{44-62i}{34}\] \[\therefore \] Real part of \[\left( \frac{{{{\bar{z}}}_{2}}{{z}_{1}}}{{{z}_{2}}} \right)=\frac{44}{34}=\frac{22}{17}\]
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