A) \[\frac{-\pi }{2}\]and 1
B) \[\frac{\pi }{2}\]and \[\sqrt{2}\]
C) 0 and \[\sqrt{2}\]
D) \[\frac{\pi }{2}\]and 1
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 \[r=\sqrt{2},\theta =\pi /4\] \[\therefore \] \[1+i=\sqrt{2}\left( \cos \frac{\pi }{4}+i\sin \frac{\pi }{4} \right)\] \[\Rightarrow \,\] \[\frac{1}{2}\,{{(1+i)}^{2}}=\frac{1}{2}\,.\,2\,{{\left( \cos \frac{\pi }{4}+i\,\sin \frac{\pi }{4} \right)}^{2}}\] By De Moivre's Theorem, \[\left( \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right)\] Hence the amplitude is \[\frac{\pi }{2}\] and modulus is 1. Trick: \[arg\text{ }\left( \frac{1+i}{1-i} \right)=arg(1+i)-arg(1-i)\] \[={{45}^{o}}-(-{{45}^{o}})={{90}^{o}}\] \[\left| \,\frac{1+i}{1-i}\, \right|\,=\frac{\left| \,1+i\, \right|}{\left| \,1-i\, \right|}\,=\frac{\sqrt{2}}{\sqrt{2}}=1\].
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