Complex numbers can be simple

Question

If !!(a+ib)^2011=a-ib!! How many ordered pairs !!(a, b)!! satisfy this equation?

  • 1
  • 4
  • 2013
  • Infinite

Solution

This is pretty straight forward question.

First we take modulus on both sides of the equation. Try to figure the reason this step.

So we get !!(a^2+b^2)^(2011/2)=(a^2+b^2)^(1/2)!!. So either !!(a^2+b^2)=0 or (a^2+b^2)=1!!

Which implies that !!a=0,b=0!! or !!(a^2+b^2)=1!!

If !!(a^2+b^2)=1=>a+ib=e^(itheta)!!

So we get !!a-ib=e^-(itheta)!!

!!(e^(itheta))^2011=e^-(itheta)!!

!!=>(e^(itheta))^2012=1!!

The solutions of the above equation are !!e^(i((2pik)/2012))!! for !!0<=k<2012!!.

So !!a+ib!! can be !!0!! or !!e^(i((2pik)/2012))!! for !!0<=k<2012!!.

Hence there are 2013 solutions

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