# ?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

Answers (1)

Solution:

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

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

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