A) \[\sqrt{2}\]
B) \[\sqrt{34}\]
C) \[1\]
D) \[\frac{2}{\sqrt{34}}\]
Correct Answer: C
Solution :
Given, arg \[\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=\frac{\pi }{2}\] \[\Rightarrow \] \[\arg \,\,{{z}_{1}}-\arg \,{{z}_{2}}=\frac{\pi }{2}\] \[\Rightarrow \] \[{{\theta }_{1}}-{{\theta }_{2}}=\frac{\pi }{2}\] Now, \[\left| \frac{3{{z}_{1}}-5{{z}_{2}}}{3{{z}_{1}}+5{{z}_{2}}} \right|\] \[=\left| \frac{3\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)-5}{3\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)+5} \right|\] \[=\left| \frac{3\,\frac{{{r}_{1}}\,{{e}^{i{{\theta }_{1}}}}}{{{r}_{2}}\,{{e}^{i{{\theta }_{2}}}}}-5}{3\,\frac{{{r}_{1}}\,{{e}^{i{{\theta }_{1}}}}}{{{r}_{2}}\,{{e}^{i{{\theta }_{2}}}}}+5} \right|\] \[=\left| \frac{3\,{{r}_{1}}\,{{e}^{i\,({{\theta }_{1}}-{{\theta }_{2}})}}-5{{r}_{2}}}{3\,{{r}_{1}}\,{{e}^{i\,({{\theta }_{1}}-{{\theta }_{2}})}}+5{{r}_{2}}} \right|\] \[=\left| \frac{3{{r}_{1}}\,i-5{{r}_{2}}}{3\,{{r}_{1}}\,i+5{{r}_{2}}} \right|\] \[\left[ \because \,\,\left( {{\theta }_{1}}-{{\theta }_{2}}=\frac{\pi }{2} \right)\,and\,{{e}^{^{i}\,\pi /2}}=i \right]\] \[=\left| \frac{-5{{r}_{2}}+3{{r}_{1}}i}{5{{r}_{2}}+3{{r}_{1}}i}\times \frac{5{{r}_{2}}-3{{r}_{1}}i}{5{{r}_{2}}-3{{r}_{1}}i} \right|\] \[=\left| \frac{-25r_{2}^{2}+9r_{1}^{2}+30{{r}_{1}}{{r}_{2}}i}{25r_{2}^{2}+9r_{1}^{2}} \right|\] \[=\sqrt{\frac{{{(-25r_{2}^{2}+9r_{1}^{2})}^{2}}+{{(30{{r}_{1}}{{r}_{2}})}^{2}}}{{{(25r_{2}^{2}+9r_{1}^{2})}^{2}}}}\] \[=\sqrt{\frac{625r_{2}^{4}+81r_{1}^{4}-450r_{1}^{2}r_{2}^{2}+900r_{1}^{2}r_{2}^{2}}{25r_{2}^{2}+9r_{1}^{2}}}\] \[=\sqrt{\frac{625r_{2}^{4}+81r_{1}^{4}+450r_{1}^{2}r_{2}^{2}}{25r_{2}^{2}+9r_{1}^{2}}}\] \[=\frac{\sqrt{{{(25r_{2}^{2}+9r_{1}^{2})}^{2}}}}{25r_{2}^{2}+9r_{1}^{2}}\]
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