Complex Numbers And Quadratic Equations question 137
Question: The maximum distance from the origin of coordinates to the point $ z $ satisfying the equation $ | z+\frac{1}{z} |=a $ is
Options:
A) $ \frac{1}{2}(\sqrt{a^{2}+1}+a) $
B) $ \frac{1}{2}(\sqrt{a^{2}+2}+a) $
C) $ \frac{1}{2}(\sqrt{a^{2}+4}+a) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ z=r $ $ (\cos \theta +i\sin \theta ) $ . Then $ | z+\frac{1}{z} |=a\Rightarrow {{| z+\frac{1}{z} |}^{2}}=a^{2} $
Þ $ r^{2}+\frac{1}{r^{2}}+2\cos 2\theta =a^{2} $ ??(i) Differentiating w.r.t. $ \theta $ we get $ 2r\frac{dr}{d\theta }-\frac{2}{r^{3}}\frac{dr}{d\theta }-4 $ sin $ 2\theta =0 $ Putting $ \frac{dr}{d\theta }=0, $ we get $ \theta =0,\frac{\pi }{2} $ $ r $ is maximum for $ \theta =\frac{\pi }{2}, $ therefore from (i) $ r^{2}+\frac{1}{r^{2}}-2=a^{2}\Rightarrow r-\frac{1}{r}=a\Rightarrow r=\frac{a+\sqrt{a^{2}+4}}{2} $
BETA
