Calculating the Galois cohomology of $U(n)$ I would like to know know what the Galois cohomology $H^1(Gal( \Bbb{C}/\Bbb{R}),U(n)(\Bbb{C}) )   $  is.
The unitary group $U(n)$ can be defined as the real form of $Gl_{n,\Bbb{C}}$ relative to the involution $\phi$ sending $A\mapsto (\bar{A}^\top)^{-1}$. Then explicitily
$$H^1(Gal( \Bbb{C}/\Bbb{R}),U(n)(\Bbb{C}) )=\{g\in GL( \Bbb{C}^n)\mid g \phi(g)=I\} /\sim $$
where the equivalence relation is induced by $ g\sim Ag\phi(A)^{-1}$ for $A\in GL(\Bbb{C}^n)$. See here for the exact definition of the first non-abelian cohomology group.
The case $n=1$ is easy and shows that the Galois cohomology is trivial, but I assume it is non-trivial for higher $n$.
 A: $\newcommand{\N}{\mathbb{N}}$$\newcommand{\Res}{\mathrm{Res}}$$\newcommand{\ad}{\mathrm{ad}}$$\newcommand{\bb}[1]{\mathbb{#1}}$$\newcommand{\C}{\mathbb{C}}$$\newcommand{\R}{\mathbb{R}}$$\newcommand{\Gal}{\mathrm{Gal}}$$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\ov}[1]{\overline{#1}}$$\newcommand{\GL}{\mathrm{GL}}$OK, so I think that I've sorted through the details now and realized that everything was easier than I thought. I am going to supress some of them here. Let me know if you want to see them.
Beore we start, let us make the convention that if $G$ is an algebraic group over a field $F$ then we shall write $H^i(F,G)$ as shorthand for $H^i(\Gal(\ov{F}/F),G(\ov{F})$ (of course $\ov{F}$ means the separable closure if we're in characteristic $p$).
Let us begin by clarifying that technically $U(n)$ is ambiguous. Namely, over $\R$ there are many groups which are called 'unitary groups'. In some sense, a 'unitary group' should just mean any form (inner or outer of $\mathrm{GL}_n$). So, to be maximally general, let's handle the case of general 'unitary groups' (i.e. all forms of $\mathrm{GL}_{n,\R}$).
Now, the forms of $\mathrm{GL}_{n,\R}$ are quite simple. They are either the inner forms of $\mathrm{GL}_{n,\R}$ or the inner forms of the group $U_{\C/\R}(n)^\ast$ which is defined to be the unitary group associated to the hermitian form $\langle -,-\rangle_0$ on $\C^n$ given by $\langle v,w\rangle:=\ov{v}^\top A w$ where $A$ is the anti-diagonal matrix with alternating anti-diagonal entries $1,-1,1,-1,\ldots$. 
(Remark: The reason that we chose these two groups $\mathrm{GL}_{n,\R}$ and $U_{\C/\R}^\ast(n)$ is that they are the only two so-called quasi-split forms of $\mathrm{GL}_{n,\R}$ which gives them special place as 'base points' (amongst inner forms) of forms of $\mathrm{GL}_{n,\R}$.)
The inner forms of $\GL_{n,\R}$ are very sparse. Namely, recall that such things are classified by $H^1(\R,\mathrm{PGL}_n)$ and this embeds into $H^2(\R,\bb{G}_m)=\mathrm{Br}(\R)[n]$. But, $\mathrm{Br}(\R)\cong \Z/2\Z$ with the only non-trivial class given by that of the Hamiltonian quaternions $\bb{H}$. From this it's not hard to see that $\GL_{n,\R}$ has no inner forms if $n$ is odd, and if $n$ is even it has only the non-trivial inner forms $\GL_{\frac{n}{2}}(\bb{H})$.
For $U_{\C/\R}(n)^\ast$ the inner forms are more rich. Namely, one can show that they are all isomorphic to $U(p,q)$ for some pair $(p,q)$ of integers with $p\geqslant q\geqslant 0$ and $p+q=n$. By $U(p,q)$ we mean the unitary group of the Hermitian space relative to $\C/\R$ with signature $(p,q)$--I trust you know what I mean by this. I assume then by $U(n)$ what you mean is $U(n,0)$ in the above notation--the anisotropic (i.e. contains no split torus) inner form of $U_{\C/\R}(n)^\ast$ (which is itsef $U(p,q)$ with $p$ as small as possible given the constraints).
That said, for Galois cohomology this is all sort of actually irrelvant given the following:

Observation: Let $k$ be a field and let $G$ and $H$ be algebraic groups over $k$ which are inner forms. Then, to any choice of inner twist $\psi:G_{\ov{k}}\to H_{\ov{k}}$ there is a naturally associated bijection $H^1(k,G)\xrightarrow{\approx}H^1(k,H)$.

Proof: See the lemma on page 7 of this. $\blacksquare$
So, the above shows us that our computations essentially reduce to the computations of the Galois cohomology of all unitary groups over $\R$ to the computation o $H^1(\R,\GL_{n,\R})$ and $H^1(\R,U_{\C/\R}(n)^\ast)$. The former of these is trivial, by Hilbert's theorem 90 (or its non-abelian variant), and so we focus on the latter.
To compute this we begin by shortening $U_{\C/\R}(n)$ to $U(n)^\ast$ (since $\C/\R$ is clear from context) and note that we have an exact sequence
$$1\to U(1)\to U(n)^\ast\to U(n)^{\ast,\ad}\to 1$$
Here for an algebraic group $G$ we denote by $G^\ad$ its associated adjoint group $G/Z(G)$. Also, note that $U(1)$ is just the circle group, which is also just $U(1)^\ast$ (there is no need to emphasize that $U(1)$ is the quasi-split inner form since there are no inner forms since it's abelian). Thus, we get an exact sequence of pointed sets
$$H^1(\R,U(1))\to H^1(\R,U(n)^\ast)\to H^1(\R,U(n)^{\ast,\ad})\to H^2(\R,U(1))\qquad(1)$$
Let us begin by computing $H^i(\R,U(1))$ for $i=1,2$. To do this let us begin by noting that we have a short exact sequence
$$1\to U(1)\to \Res_{\C/\R}\bb{G}_{m,\C}\xrightarrow{\mathrm{Nm}} \bb{G}_{m,\R}\to 1$$
where $\Res_{\C/\R}\bb{G}_{m,\C}$ is the Weil restriction (i.e. pushforward) of $\bb{G}_{m,\C}$ to $\R$ and $\mathrm{Nm}$ is the norm map. We then get a long exact sequence 
$$1\to U(1)(\R)\to \C^\times\to\R^\times\to H^1(\R,U(1))\to H^1(\R,\Res_{\C/\R}\bb{G}_{m,\C})\to H^1(\R,\bb{G}_{m,\R})\to\cdots$$
But, by Hilbert's theorem 90 as well as Shapiro's lemma we know that
$$H^i(\R,\Res_{\C/\R}\bb{G}_{m,\C})=H^i(\R,\bb{G}_{m,\R})=0,\qquad i>0$$
And from this we easily conclude that
$$H^i(\R,U(1))=\begin{cases}\R^\times/\mathrm{Nm}(\C^\times) & \mbox{if}\quad i=1\\ 0 & \mbox{if}\quad i>1\end{cases}$$
Of course, $\mathrm{Nm}(\C^\times)=\R^{>0}$ and thus $\R^\times/\mathrm{Nm}(\C^\times)\cong\Z/2\Z$. 
Thus, $(1)$ becomes
$$\Z/2\Z\to H^1(\R,U(n)^\ast)\to H^1(\R,U(n)^{\ast,\ad})\to 0$$
So, the map $H^1(\R,U(n)^\ast)\to H^1(\R,U(n)^{\ast,\ad})$ is surjective. Moreover we then know (e.g. see §5.7 of Serre's book on Galois cohomology) that there is a natural $\Z/2\Z$ action on $H^1(\R,U(n)^\ast)$ whose orbit space can be identified with $H^1(\R,U(n)^{\ast,\ad}$. Moreover, all of the orbits can be identified with a quotient of $H^1(\R,U(1))$.
Now, note that $H^1(\R,U(n)^{\ast,\ad})$ can be identified with the pointed set of (isomorphism classes) of inner forms (really inner twists, but let's ignore this important/subtle point here) and thus, by what we said above, can be identified as follows
$$\begin{aligned}H^1(\R,U(n)^{\ast,\ad}) &= \{U(p,q):p\geqslant q\geqslant 0,\,\, p+q=n\}\\ &= \{(p,q)\in \N^2:p\geqslant q,\,\, p+q=n\}\\ &= \{(p,q)\in\N^2:p+q=n\}/\sim\end{aligned}$$
where in this last term we have that $(p,q)\sim (q,p)$. 
One can then put all of the above together to identify (once one checks that at most one quotient of $H^1(\R,U(1))$ by $H^0(\R,{_\psi}U(n)^{\ast,\ad}(\C))$ is non-trivial where $\psi$ is an element of $H^1(\R,U(n)^\ast))$) that
$$H^1(\R,U(n)^\ast)=\{(p,q)\in\N^2:p+q=n\}$$
and the projection map 
$$H^1(\R,U(n)^\ast)\to H^1(\R,U(n)^{\ast,\ad})$$
can be identified with the quotient map
$$\{(p,q)\in\N^2:p+q=n\}\to \{(p,q)\in\N^2:p+q=n\}/\sim$$
Hopefully this is the kind of answer you wanted. In particular, we see that $\# H^1(\R,U(n)^\ast)=n+1$.

I wanted to add that I had occasion to reread portions of the papers of Borovoi and realized that while the one theorem of Borovoi-Kottwitz (that the abelianized cohomology agrees with cohomology over a $p$-adic local field) doesn't work in $\mathbb{R}$ setting there is an even nicer result in the $\mathbb{R}$ setting contained in another paper of Borovoi!
Namely, there is the following nice theorem (I don't know who it is occasionally due to--versions of it are due to Serre and Cartan) which makes the whole discussion largely combinatorial:

Theorem: Let $G$ be a reductive group over $\mathbb{R}$ and let $T$ be a fundamental torus of $G$ (i.e. one of minimal split rank). Then, the natural map 
  $$H^1(\mathbb{R},T)\to H^1(\mathbb{R},G)$$
  is surjective. Moreover, if we let $W_T$ be the Weyl group scheme, then $W_T(\mathbb{R})$ naturally acts on $H^1(\mathbb{R},T)$ in way equivariant for for the projection to $H^1(\mathbb{R},G)$. Moreover, the induced map
  $$H^1(\mathbb{R},T)/W_T(\mathbb{R})\to H^1(\mathbb{R},G)$$
  is a bijection.

Proof: This is Theorem 9 of this. $\blacksquare$
Let us apply this to the case when $G=U(n)=U(n,0)$ (again, you can replace $G=U(n)$ with any inner forms, like $U(n)^\ast$, but it doesn't matter because of the observation we made). As a fun exercise, first show this

Lemma: Let $G$ be a reductive group over $\mathbb{R}$ and suppose that $T$ is an elliptic maximal torus (i.e. $T/Z(G)$ is $\mathbb{R}$-anisotropic). Then, the group scheme $W_T$ is constant.

In particular, note that $T=U(1)^n$ is an $\mathbb{R}$-anistropic, thus a fortiori an elliptic, maximal torus in $G=U(n)$. Note then that by our above calculations
$$H^1(\mathbb{R},T)=H^1(\mathbb{R},U(1)^n)=H^1(\mathbb{R},U(1))^n=(\mathbb{Z}/2\mathbb{Z})^n$$
Now by our lemma we have that $W_T$ is a constant group scheme. But, evidently $W_T(\mathbb{C})$, being the Weyl group of a maximal torus in $\mathrm{GL}_{n,\mathbb{C}}$, is $S_n$. Thus, by constancy $W_T(\mathbb{R})=S_n$. It's not hard to see that $S_n$ acts on $H^1(\mathbb{R},U(1)^n)=(\mathbb{Z}/2\mathbb{Z})^n$ by permuting factors the factors. Thus, we see that by the above theorem that 
$$H^1(\mathbb{R},U(n))\cong (\mathbb{Z}/2\mathbb{Z})^n/S_n\cong\{(p,q)\in\mathbb{N}^2:p+q=n\}$$
Repeating this argument with $G=U(n)^\mathrm{ad}$ shows that we can identify 
$$H^1(\mathbb{R},U(n)^\mathrm{ad})=((\mathbb{Z}/2\mathbb{Z})^n/(\mathbb{Z}/2\mathbb{Z})/S_n\cong \{(p,q)\in\mathbb{N}^2:p+q=n\}/\sim$$
where $\mathbb{Z}/2\mathbb{Z}$ is embedded diagonally. And then it's not hard to show by functoriality that the map
$$H^1(\mathbb{R},U(n))\to H^1(\mathbb{R},U(n)^\mathrm{ad}$$
is the obvious quotient map as in the previous version of the answer. Moreover, it's easy to see that the diagonal copy of $\mathbb{Z}/2\mathbb{Z}$ in this version of the answer is accounted for by $H^1(\mathbb{R},U(1))$, where $U(1)$ is embedded centrally in $U(n)$, from the previous version of the answer. 
