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