A) \[1\]and \[\frac{\pi }{4}\]
B) \[\sqrt{2}\]and\[\frac{\pi }{3}\]
C) 1 and \[\frac{\pi }{2}\]
D) \[-1\]and \[-\pi /2\]
Correct Answer: C
Solution :
\[z=\frac{1+i}{1-i}\times \frac{1+i}{1+i}=\frac{{{(1+i)}^{2}}}{1-{{i}^{2}}}=\frac{1+{{i}^{2}}+2i}{2}=i\] \[i=r(\cos \theta +i\sin \theta )\] \[\Rightarrow \] \[r\cos \theta =0,\,1=r\sin \theta \] \[{{r}^{2}}{{\cos }^{2}}\theta =0,{{r}^{2}}{{\sin }^{2}}\theta =1\] \[\Rightarrow \] \[{{r}^{2}}=1\] \[\Rightarrow r=1\] \[\tan \theta =\frac{1}{0}=\infty \Rightarrow \theta =\pi /2\]
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