A) \[\sqrt{2}\text{ and }\frac{\pi }{6}\]
B) 1 and 0
C) 1 and \[\frac{\pi }{3}\]
D) 1 and \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
\[\frac{1+2i}{1-{{(1-i)}^{2}}}=\frac{1+2i}{1-(1-1-2i)}=\frac{1+2i}{1+2i}=1+0i\] Modulus =1 Amplitude \[\theta \]=\[{{\tan }^{-1}}\frac{0}{1}=0\].You need to login to perform this action.
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