Complex Numbers And Quadratic Equations question 85
Question: Argument and modulus of $ \frac{1+i}{1-i} $ are respectively [RPET 1984; MP PET 1987; Karnataka CET 2001]
Options:
A) $ \frac{-\pi }{2} $ and 1
B) $ \frac{\pi }{2} $ and $ \sqrt{2} $
C) 0 and $ \sqrt{2} $
D) $ \frac{\pi }{2} $ and 1
Show Answer
Answer:
Correct Answer: D
Solution:
$ \frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{{{(1+i)}^{2}}}{2} $ Now $ 1+i=r(\cos \theta +i\sin \theta )\Rightarrow r\cos \theta =1,r\sin \theta =1 $
Þ $ r=\sqrt{2},\theta =\pi /4 $
$ \therefore $ $ 1+i=\sqrt{2}( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} ) $
$ \Rightarrow $ $ \frac{1}{2}{{(1+i)}^{2}}=\frac{1}{2}.2{{( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} )}^{2}} $ By De Moivre’s Theorem, $ ( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} ) $ Hence the amplitude is $ \frac{\pi }{2} $ and modulus is 1. Trick: $ arg( \frac{1+i}{1-i} )=arg(1+i)-arg(1-i) $ $ =45^{o}-(-45^{o})=90^{o} $ $ | \frac{1+i}{1-i} |=\frac{| 1+i |}{| 1-i |}=\frac{\sqrt{2}}{\sqrt{2}}=1 $ .
BETA
