why is $e^{\frac{\pi i}{3}}$ a root of $f(x) = x^{2}-x+1$?

Question

I am not sure how this comes about since, $f(e^{\frac{\pi i}{3}})=(e^{2 \pi i})^{\frac{1}{3}}-(e^{\pi i})^{\frac{1}{3}}+1 = (1)^{\frac{1}{3}}-(-1)^{\frac{1}{3}}+1 = 1+1+1 = 3$

Answer 1

Multiply $f(x)=x^2-x+1$ with $x+1$ to get $g(x)=(x+1)f(x) = x^3+1$ which means the zeros of $g$ are the 3rd roots of $-1$. Thus the roots of $f$ are the 3rd roots of $-1$ but not $-1$ itself. Now $(e^{\pi i/3})^3 = e^{\pi i}=-1$ and also $e^{\pi i/3}\neq1$.

Answer 2

In $\mathbb C$, the equality $(e^a)^b=e^{ab}$ doesn't make sense in general and, even when it does, it is not necessarily true (although it works when $b\in\mathbb Z$). But you used in your computations.

On the other hand,\begin{align}\require{cancel}f(e^{\pi i/3})&=(e^{\pi i/3})^2-e^{\pi i/3}+1\\&=e^{2\pi i/3}-e^{\pi i/3}+1\\&=\cos\left(\frac{2\pi}3\right)+\sin\left(\frac{2\pi}3\right)i-\cos\left(\frac\pi3\right)-\sin\left(\frac\pi3\right)i+1\\&=\bcancel{-\frac12}+\cancel{\frac{\sqrt3}2i}\bcancel{-\frac12}-\cancel{\frac{\sqrt3}2i}+\bcancel1\\&=0.\end{align}

Answer 3

$x=e^{i\pi/3}=1/2+\sqrt{3}i/2.$ Then $$f(e^{i\pi/3})=[1/2+i\sqrt{3}/2]^2-[1/2+i\sqrt{3}/2]+1=1/4-3/4+i\sqrt{3}/2-1/2-i\sqrt{3}/2+1=0$$

Answer 4

$(z^a)^b \ne (z^b)^a$ if $a,b$ aren't both integers.

Indeed $z^{\frac 1n}$ is not well (or usefull) defined at all. There are $n$ values of $w$ so that $w^3 = z$ so if we decide arbitrarily that one is correct and the other $n-1$ aren't we can do anything because if we have $w^n = z$ we can't play God and assume that one of them is right and the others aren't.

$e^{\frac {2\pi i}n} := \cos \frac {2\pi}n + i\sin \frac {2\pi}n$.

It is NOT true that $e^{\frac{2\pi i}n} = (e^{2\pi i})^{\frac 13}$. And although $e^{2\pi i} =1$. And $1^{\frac 13}$ is not well defined and it doesn't make any sense to say "$1^{\frac 13} =1$" when talking of complex numbers.

$f(e^{\frac {2\pi i}3}) = (e^{\frac {2\pi i}3})^2 -e^{\frac {2\pi i}3} + 1=$

$e^{\frac {4\pi i}3}-e^{\frac {2\pi i}3} + 1=$

$(\cos \frac {4\pi}3 + i\sin \frac {4\pi}3) - (\cos \frac {2\pi}3 + i\sin \frac {2\pi}3) + 1=$

$(-\frac 12-i\frac {\sqrt 3}2)-(-\frac 12+i\frac {\sqrt 3}2) + 1 = $

$(-\frac 12 -\frac 12) + (-\frac {\sqrt 3}2 + \frac {\sqrt 3}2)i + 1=$

$-1 + 0i + 1 = 0$.

Answer 5

Attempt:

$x^3+1=(x+1)(x^2-x+1)$;

$x^3=-1$ has roots

$x_1=e^{πi/3}; x_2=e^{2πi/3}$;

$x_3=e^{πi}=-1,$ the real root.

$x_2,x_3$ are the roots of $x^2-x+1$.