symplectic structure of tangent bundle of $\mathbb{S}^{n-1}$ It is  well known that   $T\mathbb{S}^{n-1}$ is diffeomorphic to $M= f^{-1}(1)$ where 
$f:\mathbb{C}^n\rightarrow \mathbb{C}$ is $f(z):=\sum_{i=1}^{n} z_{i}^{2}$.
Two questions:
1) Is  $M$  a  symplectic  submanifold of $\mathbb{C}^n\sim \mathbb{R}^{2n}$ (with the standard symplectic structure)?
If  the answer is affirmative, we can consider two  symplectic structures for $T\mathbb{S}^{n-1}$. The first is the original structure of the tangent or  cotangent bundle, the second one is the  pull back  structure of  $M$.
2) Are these structures equivalent? 
 A: $
\def\C{{\mathbf C}}
\def\d{\delta}
\def\e{{\mathbf e}}
\def\r{{\mathbf r}}
\def\u{{\mathbf u}}
\def\x{{\mathbf x}}
\def\y{{\mathbf y}}
\def\z{{\mathbf z}}
\def\<{\langle}
\def\>{\rangle}
$
If I understand Now that I understand your questions correctly, the answers are 1) Yes 2) No Yes.
1) Writing $\z=\x+i\y$, we have $f(\z) = \|\x\|^2 - \|\y\|^2+2i\<\x,\y\>$. So $M=f^{-1}(1)$ (resp. $\mathrm T_\z M$) consists of all $\z$ (resp. $\d\z=\d\x+i\d\y$) such that
$$
\|\x\|^2 - \|\y\|^2= 1, \qquad \<\x,\y\>=0,
\tag1
$$
$$
\llap{(\text{resp.}\qquad}\<\x,\d\x\>-\<\y,\d\y\>=0,\qquad\<\d\x,\y\>+\<\x,\d\y\>=0.)
\tag2
$$
Given $\z$ in (1), we can choose an orthonormal basis $(\e_1,\dots,\e_n)$ such that $\x = \|\x\|\e_1$ and $\mathbf y = \|\y\|\e_2$. Then (2) says that $(\d x_1,\d y_1)=\frac{\|\y\|}{\|\x\|}(\d y_2,-\d x_2)$ and hence
\begin{align}
\omega(\d\z,\d'\z)
&=
\sum_{i=1}^n(\d x_i\d'y_i-\d'x_i\d y_i)\\
&=\Bigl(1+\tfrac{\|\y\|^2}{\|\x\|^2}\Bigr)(\d x_2\d'y_2-\d'x_2\d y_2)
+ \sum_{i=3}^n(\d x_i\d'y_i-\d'x_i\d y_i)
\end{align}
which is obviously nondegenerate. (Or as Tim Perutz says, just note that $M$ is a complex submanifold.)
2) Consider (after Seidel's thesis pointed out by Daniel Pomerleano) the diffeo $\Phi:M\to\mathrm{TS}^{n-1}$ sending $\z=\x+i\y$ to the pair $(\u,\r)=(\frac{\x}{\|\x\|}, \|\x\|\y)$. Writing $\sigma$ for the canonical 2-form $d\<\u,d\r\>$ on $\mathrm{TS}^{n-1}$, we have
\begin{align}
\Phi^*\sigma &= d(\Phi^*\<\u,d\r\>)=d(\<\tfrac{\x}{\|\x\|},d(\|\x\|\y)\>)\\
&=d\<\x,d\y\> + \<\x,\y\>\tfrac{d\|\x\|}{\|\x\|}
\end{align}
and this equals $\omega$ since the second term vanishes by (1). Thus $\Phi$ is a symplectomorphism.
3) My initial (stupid and wrong) answer came from my thinking of another similar question. Namely $\mathrm{TS}^2$, thought of as the manifold of oriented affine lines in $\mathbf R^3$ (identify $(\u,\r)$ with the line $\r+\mathbf R\u$ with orientation $+\u$), carries a 2-parameter family of Euclid-invariant symplectic structures given by
$$
\omega_{k,s} = d\<k\u,d\r\> - s\Omega,\qquad k>0, s\in \mathbf R,
\tag3
$$
where the first term is the cotangent bundle 2-form, and the second ("magnetic") term is $-s$ times the area 2-form of the base $\mathrm S^2$.
On the other hand, identifying $\mathfrak{so}(3,\C)=\C^3$ with Lie bracket the vector product, the nonzero levels of your quadratic form $\<\z,\z\>=\sum_{i=1}^3 z_i^2$ are precisely the semisimple (co)adjoint orbits of $\mathrm{SO}(3,\C)$. As such they carry a (Kirillov-Kostant-Souriau) 2-form $\omega_{\mathrm{KKS}}(\d\z,\d'\z)=\mathrm{Re}\bigl(\frac1{s^2}\<\z,\d'\z\times\d\z\>\bigr)$ where $\<\z,\z\>=s^2$ is the equation of our orbit $\mathcal O$. So we can again ask if this is symplectomorphic with $\mathrm{TS}^2$ equipped with one of the forms (3). At least when $s^2$ is real positive, the answer is yes and the diffeomorphism $F:\mathrm{TS}^2\to\mathcal O$ I was considering (with $k=s=1$),
$$
F(\u,\r) = k\r\times\u + s\u + ik\r
$$
is symplectic: $F^*\omega_{\mathrm{KKS}}=\omega_{k,s}$. Being familiar with this diffeomorphism, I then made the mistake of thinking it must be "the one" of which you were asking if it was symplectic relative to your (different) 2-forms. And it isn't.
A: Here is a formula for an explicit symplectomorphism $F$ from $T^*S^{n-1}$ to the affine quadric $\{\sum z_{j}^{2}=1\}$ in $\mathbb{C}^n$: 
$$ 
F(p,q) = \left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{1/2} q - i\left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{-1/2}p 
$$
Here I view $T^*S^{n-1}$ as consisting of pairs $(p,q)\in \mathbb{R}^n\times\mathbb{R}^n$ with $p\cdot q=0$ and $|q|=1$.  To check that $F$ is a symplectomorphism one can just check that $F$ pulls back the one-form $\alpha=-\sum y_jdx_j$ on $\mathbb{C}^n$ (which is a primitive for the standard symplectic form) to the canonical one-form on $T^*S^{n-1}$.  The coefficients involving $\sqrt{1+4|p|^2}$ are designed to cause the map to have image in the quadric.  
To indicate how I came up with this formula, the reasoning was that a suitably natural symplectomorphism ought to be equivariant with respect to the natural Hamiltonian $O(n)$-actions on both spaces.  The easiest way to do this seems to be to take $F(p,q)=f(|p|)q-ig(|p|)p$ for some real functions $f$ and $g$.  For the image to be in the quadric one needs $$f(|p|)^2- |p|^2g(|p|)^2=1.$$  Meanwhile if this is to be a symplectomorphism it should pull back the moment map for the $O(n)$-action on $\mathbb{C}^n$ to the moment map for the $O(n)$-action on $T^*S^{n-1}$.  The norms of these moment maps are, respectively $(x+iy)\mapsto |x||y|$ and $(p,q)\mapsto |p|$, giving an equation $|f||g||p|=|p|$, i.e., $$|g|=\frac{1}{|f|}.$$ Solving these for $f$ and $g$ yields the formula at the top, which can then be directly confirmed to have the required properties.
