# Complex Numbers problem

1. Apr 30, 2014

### sankarshana016

If a is a complex number, and a^2-a+1=0, then a^2011=?

I tried using De Moivre's theorem, Taking a=cosθ+isinθ, but didn't get anywhere, got stuck at
cos2θ+isin2θ-cosθ-isinθ+1=0. What do I do?

2. Apr 30, 2014

### csopi

First you should solve the equation for 'a'. After that, you can calculate the argumentum and the abs. value of 'a', resulting in a = r*(cos(theta) + i* sin(theta)). Finally, apply De Moivre for calculating a^2011.

3. Apr 30, 2014

### AlephZero

a=cosθ+isinθ assumes that |a| = 1. That is wrong unless you already know that |a| = 1.

You can just solve the quadratic equation for a (using the standard formula) to get the real and imaginary parts.

4. Apr 30, 2014

### SammyS

Staff Emeritus
Do you know the factorization of $\ a^3+1\ ?$

5. Apr 30, 2014

### Dick

It might be difficult to guess that clever trick. Here's another way. You know a^2=a-1. Can you find a simple expression for a^3?

6. May 1, 2014

### sankalpmittal

Dude, first find the complex number "a" by using quadratic formula. Then you convert it to "cis" notation, and then you finally apply De Moivre's theorem.

7. May 3, 2014

### lurflurf

$a^{2011}=a^{6\cdot 335+1}=(a^6)^{335}a$
What is $a^6$?