Complex Numbers 1.02

Algebra Level 5

Number of ordered points $(a,b)$ of real numbers such that $(a+ib)^{2011}=a-ib$ is

2010 2012 2011 2013

2 solutions

Anirudha Nayak
Mar 25, 2014

Take modulus on both side........

The answer is 2013, not 2012. One solution is $(0,0)$ , and there are 2012 solutions that correspond to the 2012nd roots of unity. See https://www.artofproblemsolving.com/Wiki/index.php/2002 AMC 12A Problems/Problem 24 for a related problem.

Jon Haussmann - 7 years, 2 months ago

Then what?????

Kaushik Aranagu - 7 years, 2 months ago

i'l post the solution on 29th try using C.N property

Anirudha Nayak - 7 years, 2 months ago

You have to be more specific than that. Could you reiterate algebraically?

Julien Bongars - 6 years, 8 months ago

Aniket Sanghi
Feb 10, 2017

Take mod on both side will yield |z| = 0 or 1.

Case 1 : Take |z| = 1. Let $z = 1 × {e}^{i\theta}$

Then we see ${e}^{2011i\theta} = {e}^{-i\theta}$

Implies $2011\theta - 2n \pi = - \theta$ (n is an integer.) ( As Principle value )

You will get values for $n \quad belongs \quad to \quad ( -1006 , 1006]$

That is 2012 solution

Case 2 : |z| = 0. This also satisfies equation ...

Hence 2013 solutions ...