The Whitney sphere immersion The standard unit sphere $S^n = \{ (x,y) \in \mathbb R^n \times \mathbb R  \mid \lvert x \rvert^2 + y^2  = 1\}$ can be immersed inside $\mathbb C^n$ through the formula $(x,y) \mapsto (1+iy)x$. This is a Lagrangian immersion with a double point at the poles $(0,\pm 1)$. In the special case $n = 1$, this yields a figure 8 in the plane.

What is an easy way to see that this formula does indeed yield an immersion?

The case $n = 1$ is easy enough, but things get ugly for arbitrary $n$. I've tried using generalized spherical coordinates, but the matrix of the differential in these coordinates is nasty, and it's unclear to me why it has maximal rank. Alternatively, viewed as a map $\mathbb R^{n+1} \to \mathbb R^{2n}$, the immersion has a nice enough differential, but these rectangular coordinates are not adapted to the tangent bundle of the sphere.
 A: It is easier to split into two cases. 
When $y\neq 0$, the set $\mathbb S^n_\pm = \{((x, y)\in \mathbb S^n: \pm y>0\}$ admits charts $\phi_\pm:D(1) \to \mathbb S^n$
$$\phi_{\pm} (x)= (x, \pm \sqrt{1-|x|^2}) $$
Under this chart, the immersion $W(x,y)= (1+iy)x$ is 
$$ W\circ \phi_\pm (x) = (1\pm i\sqrt{1-|x|^2}) x = x \pm i x \sqrt{1-|x|^2},$$
which is clearly an immersion since it is a graph $x\mapsto (x, f(x))$ (treating $\mathbb C^n = \mathbb R^n \oplus \mathbb R^n$ here). Note that from here we also show that $W$ is Lagrangian as 
$$\pm x\sqrt{1-|x|^2} = \pm \frac 13\nabla (1-|x|^2)^{3/2}.$$
When $y = 0$, $W(x, 0) = x$. It is very clear that it is an immersion when $W$ is restricted to the equator $\{(x, y): y=0\}$. So it suffices to check $(W_*)_{(x,0)} (0,\cdots, 0, 1)$. By definition of the differential,
$$\begin{split}
(W_*)_{(x,0)} (0,\cdots, 0, 1)&= \frac{d}{dt}\bigg|_{t=0} W((\cos t)x, \sin t ) \\
&= \frac{d}{dt}\bigg|_{t=0} (1+ i\sin t) (\cos t) x \\
&=  ix.
\end{split}$$
Note that this implies $(W_*)_{(x,0)}$ is also injective as $i x$ lies outside the real $n$ plane $\mathbb R^n \oplus \{0\}$ (where all others $(W_*)_{(x,0)} (v)$ lives for $v$ in the tangent bundle of the equator)
A: Well, I feel a bit silly. One actually can use the differential of the map $\mathbb{R}^{n+1} \to \mathbb{R}^{2n}$ to conclude. This differential at the point $(x,y) \in S^n \subset \mathbb R^{n+1}$ is just given by $(u,v) \mapsto (u,yu+vx) \in \mathbb R^{2n}$ ($u \in \mathbb R^n$, $v \in \mathbb R$). If this is zero, then $u = 0$ and $vx = 0$, and if $x = 0$ then we are at one of the poles $(0,\pm 1)$, and we must have $v=0$ to have a tangent vector there. Therefore the differential is injective.
