Complex Numbers And Quadratic Equations question 74
Question: Let z be a complex number (not lying on X-axis of maximum modulus such that $ | z+\frac{1}{z} |=1 $ . Then
Options:
A) $ Im(z)=0 $
B) $ Re(z)=0 $
C) $ amp(z)=\pi $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ z=r(\cos \theta +i\sin \theta ) $ . Then $ | z+\frac{1}{z} |=1\Rightarrow {{| z+\frac{1}{z} |}^{2}} $ =1
Þ $ {{| r(\cos \theta +i\sin \theta )+\frac{1}{r}(\cos \theta -i\sin \theta ) |}^{2}}=1 $ .
Þ $ {{( r+\frac{1}{r} )}^{2}}{{\cos }^{2}}\theta +{{( r-\frac{1}{r} )}^{2}}{{\sin }^{2}}\theta =1 $
Þ $ r^{2}+\frac{1}{r^{2}}+2\cos 2\theta =1 $ Since $ |z|=r $ is maximum, therefore $ \frac{dr}{d\theta }=0 $ Differentiating (i) 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 $ \sin 2\theta =0 $
Þ $ \theta =0 $ or $ \frac{\pi }{2} $
Þ z is purely imaginary or purely real. ( $ \because \theta =0 $ is not given)
BETA
