A) 4
B) 3
C) 2
D) 5
Correct Answer: A
Solution :
\[{{x}^{2}}-2x+2=0\] |
\[{{(x-1)}^{2}}=-1={{i}^{2}}\] |
\[x=1+i,1-i\] |
Let, \[\alpha =1+i,\beta =1-i\] |
\[{{\left( \frac{\alpha }{\beta } \right)}^{n}}=1\] \[\Rightarrow \]\[{{\left( \frac{1+i}{1-i} \right)}^{n}}=1\]\[\Rightarrow \]\[{{\left( \frac{\sqrt{2}{{e}^{i\pi /4}}}{\sqrt{2}{{e}^{-i\pi /4}}} \right)}^{n}}=1\]\[\Rightarrow \]\[{{e}^{1n\pi /2}}=1\]\[\Rightarrow \]\[n=4.\] |
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