Let me elaborate:
By properties, I mean qualities such as "is a submersion", "is an immersion", "is an embedding", etc.
Let's say I have a function $f:S^1 \rightarrow S^3$ between the spheres viewed as smooth manifolds in $\mathbb{C}$ and $\mathbb{C}^2$ respectively.
In class we saw that to prove for example that $f$ is a submersion, we should first define maps (or "charts") $\varphi:\mathbb{R^2}\rightarrow S_1$, $\psi:S^3\rightarrow \mathbb{R}^4$ and check that the differential $D_x(\psi \circ f \circ \varphi)$ is onto.
In other words, to show $f$ has one of the properties, we first have to view it as a function from $\mathbb{R}^2$ to $\mathbb{R^4}$.
I would like to know if it is equivalent to view $f$ as a function from $\mathbb{C}$ to $\mathbb{C^2}$ instead when it makes things much more convenient and simply differentiate with respect to the complex variable as if it were real since it would simplify a lot the calculations (here $\varphi$ and $\psi$ would simply be identity! or rather, inclusion from the spheres to the complex plane)
In particular, would this work for all complex functions? Or only those that are "analytic" (e.g. which do not include terms in $\bar z$)?
To give a concrete example, let's consider:
$f(z)=(cz^n,c^{m/n}e^{i\pi/n}z^m)$ where $c$ is real positive chosen such that $f$ is from $S_1$ to $S_3$ (i.e. if $|z|=1$ then $|f(z)|=1)$
Say I want to know whether it is a immersion. The "safe" way would be to convert $z=x+iy$ to $(x,y)$ in $\mathbb{R}^2$ and get a function sending $(x,y)$ to $(a,b,c,d)$ from $\mathbb{R}^2$ to $\mathbb{R}^4$ where $a,b,c,d$ are functions of $x$ and $y$. But computing this could be tedious depending on the function involved, so can I simply compute the complex differential with respect to $z$ and determine whether it is injective? (from $\mathbb{C}$ to $\mathbb{C^2}$)
For example here I would get $D_zf(h)=(cnz^{n-1},c^{m/n}e^{i\pi/n}mz^{m-1}).h$ ($h\in \mathbb{C})$
and $D_zf(h)=0 \Leftrightarrow h=0$ assuming $c>0$ and $n>0$ (and $z\neq0$ but here $z\in S_1$ so it is automatically true)
so $KerD_zf=0$ and $f$ is an immersion. Is this correct? Would this still work if there were non-analytic terms (e.g. $\bar z$ instead of $z$)?