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  • ?If a and b be the roots of the equation x^2-2x+2=0, then the least value of n for which (a/b)^n=1 is

?If a and b be the roots of the equation x^2-2x+2=0, then the least value of n for which (a/b)^n=1 is

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best_answer

 

Solution:

\ x^2-2x+2=0\ \Rightarrow x=(1pm i)

Then \(a/b)=(1+i)/(1-i)=(1+i)^2/1-i^2=i

or \(a/b)=(1-i)/(1+i)=(1-i)^2/1-i^2=-i

So ,\(a/b)=pm i

Now, \ (a/b)^n=1\ \Rightarrow (pm i)^n=1

⇒ Minimum value of n =4

Posted by

Deependra Verma

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