Question

# If $$\dfrac{5z_{1}}{7z_{2}}$$ is purely imaginary, then $$\left |\displaystyle \frac{2z_{1}+3\mathrm{z}_{2}}{2z_{1}-3\mathrm{z}_{2}} \right | is$$

A
57
B
75
C
1
D
35

Solution

## The correct option is C $$1$$$$\dfrac{5}{7}\dfrac{z_{1}}{z_{2}}$$ is purely imaginary $$\Rightarrow \dfrac{z_{1}}{z_{2}}$$ is purely imaginary$$\dfrac{z_{1}}{z_{2}}+\dfrac{\bar{z_{1}}}{\bar{z_{2}}}=0\Rightarrow z_{1}\bar{z_{2}}+z_{2}\bar{z_{1}}=0 --(1)$$Now,$$\left |\dfrac{2z_{1}+3z_{2}}{2z_{1}-3z_{2}} \right|^{2}=\dfrac{(2z_{1}+3z_{2}) \times (2\bar{z_{1}}+3\bar{z_{2}})}{(2z_{1}-3z_{2}) \times (2\bar{z_{1}}-3\bar{z_{2}})} \quad \left \{ as |z|^{2}=z.\bar{z} \right \}$$$$=\dfrac{4|z_{1}|^{2}+6z_{1}\bar{z_{2}}+6z_{2}\bar{z_{1}}+9|z_{2}^{2}|}{4|z_{1}|^{2}-6z_{1}\bar{z_{2}}-6z_{2}\bar{z_{1}}+9|z_{2}|^{2}}$$$$=\dfrac{4|z_{1}|^{2}+9|z_{2}|^{2}+6(z_{1}\bar{z_{2}}+z_{2}\bar{z_{1}})}{4|z_{1}|^{2}+9|z_{2}|^{2}-6(z_{2}\bar{z_{1}}+z_{1}\bar{z_{2}})}$$But from (1) ,$$z_{1}\bar{z_{2}}+z_{2}\bar{z_{1}}=0$$$$=\dfrac{4|z_{1}|^{2}+9|z_{2}|^{2}}{4|z_{1}|^{2}+9|z_{2}|^{2}}=1$$Mathematics

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