How to Solve a^2011 for a Complex Number Satisfying a^2-a+1=0?

[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?

 

[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.

 

[AlephZero]

Taking a=cosθ+isinθ, but didn't get anywhere

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.

 

[SammyS]

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?

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

 

[Dick]

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

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?

 

[sankalpmittal]

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?

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.

 

[lurflurf]

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