A) \[\sqrt{2}\] and \[\frac{\pi }{6}\]
B) 1 and 0
C) 1 and \[\frac{\pi }{3}\]
D) 1 and \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
Let \[z=\frac{1+2i}{1-{{(1-i)}^{2}}}\] \[=\frac{1+2i}{1-({{1}^{2}}+{{i}^{2}}-2i)}=\frac{1+2i}{1+2i}\] \[=1\] \[\therefore \,\,\,|z|=1\] and amp \[(z)={{\tan }^{-1}}\left( \frac{0}{1} \right)=0\]You need to login to perform this action.
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