Twist in identification with singular cohomology Let $X$ be a smooth projective variety over $\mathbb{Q}$ and $$V = H^m(X(\mathbb{C}), \mathbb{Q} \cdot (2\pi i)^r)$$ Then I've seen people write the comparison with complex cohomology (an isomorphism of $\mathbb{Q}_\ell$ vector spaces) as $$V \otimes_\mathbb{Q} \mathbb{Q}_\ell \cong H^m_{et}(X_{\bar{\mathbb{Q}}}, \mathbb{Q}_\ell)(r)$$
Why does twisting on the right hand side correspond to taking coefficients in $\mathbb{Q} \cdot (2\pi i)^r$ instead of $\mathbb{Q}$? What does the twist do, what is it used for?
 A: Let me supplement Vivek's nice answer with a few additional comments. People coming from differential geometry typically normalize the $n$th Chern class -- in the Chern-Weil formula -- by a factor of $1/(2\pi i)^n$ to get it to be integral. However, if you want compatibility with algebraic definitions, then you don't want to do this, so then $c_n$ would take values in $H^{2n}(X,(2\pi i)^n\mathbb{Z})$. For example, in additional to the etale  first Chern class explained above, you can use algebraic de Rham cohomology. Here $c_1$ is given by the induced map on $H^2$ associated to 
$$ d\log:\mathcal{O}_X^*[-1]\to \Omega_X^\bullet$$
Continued-- To expand slightly, one can see by a diagram chase that, under the Grothendieck isomorphism $H^2(X,\mathbb{C})\cong H^2(X,\Omega_X^\bullet)$, the image of this $c_1$ lands in $H^2(X,2\pi i\mathbb{Z})$. So by the usual tricks involving the splitting principle, one gets the a similar statement for higher Chern classes. In case, you prefer to avoid Chern classes, it follows (with some work), that the image of the cycle map $CH^n(X)_\mathbb{Q}\to H^{2n}(X, \mathbb{C})$ lands in $H^{2n}(X, \mathbb{Q}(n))$. Again this is consistent with what happens on the etale side.
A: I think at some level this is a convention having to do with the fact that our exponential is $x \mapsto e^x$ rather than $x \mapsto e^{2\pi i x}$.  But having made this choice, here's one way that factor appears:
Over $\mathbb{C}$, you have the exponential sequence 
$$0 \to 2\pi i \mathbb{Z} \to \mathbb{G}_a \to \mathbb{G}_m \to 1$$
Line bundles are classified by $H^1(\,\cdot\,, \mathbb{G}_m)$, and the first Chern class is the natural map 
$$c_1: H^1(\, \cdot \, , \mathbb{G}_m) \to H^2(\,\cdot \,, 2\pi i\mathbb{Z})$$
There's also the $\ell$-th power map.  If $\mu_\ell$ is the group of $\ell$-th roots of unity, then you get a sequence:
$$1 \to \mu_\ell \to \mathbb{G}_m \to \mathbb{G}_m \to 1$$ 
Taking etale cohomology gives a notion of Chern class sensible over any field
(... of characteristic prime to $\ell$):
$$c_1: H^1_{et}(\, \cdot \, ,\mathbb{G}_m) 
\to H^2_{et}(\,\cdot \,, \mu_\ell)$$  
The first sequence maps to the second by sending the first two terms via $\mathrm{exp}(\,\cdot\, /\ell)$, and the third by the identity.  The second sequence for $\mu_{\ell^2}$ maps to that for $\mu_\ell$ by the $\ell$-th power map on the first two terms and the identity on the third.  By definition we have $\lim \mu_{\ell^n} = \mathbb{Z}_\ell(1)$, so taking an inverse limit gives
$$c_1: H^1_{et}(\, \cdot \, ,\mathbb{G}_m) 
\to H^2_{et}(\,\cdot \,, \mathbb{Z}_\ell(1))$$  
So if you want your Chern classes to line up, it's nice to identify the Tate twist $(1)$ in etale cohomology with $\cdot 2\pi i$ in the singular theory. 
