Problems about homology and homology group In my algebraic topology class I was assigned to do the following problems
(1) Let $f:\mathbb{C}^2\to \mathbb{C}$ be a nonzero complex linear form.
Show that $\mathbb{C}^2\setminus f^{-1}(0) $ is homotopic to circle $S^1$
Somehow $\mathbb{C}^2\setminus f^{-1}(0) $ reminded me of the homomorphism theorem so this should be $\cong Im(f)$Do I misunderstand it? however I cannot imagine the image of this set so I still have no idea how to find deformation retract.Can you explain how to show this?
(2) $A=\{(x,y)\in\mathbb{C}^{2}\mid (x,y)\not= (0,0)\}$
$B=\{(x,y)\in\mathbb{C}^{2}\mid xy\not= 0\}$
then find the homology group $H_n(A)=H_n(A,\mathbb{Z})$and $H_n(B)=H_n(B,\mathbb{Z})$
when $n\in\mathbb{N}\geq 0$
last week I learnt about Mayer-Vietoris and how to find homology group of sphere so I wonder if it can be applied here. 
Thanks for your help.
 A: Here is a hint.  First, identify $\mathbb{C}^2 \cong \mathbb{R}^4$, and try to prove the following:
Lemma: Let $W$ be a $k$-dimensional linear subspace of $\mathbb{R}^n$ with $k < n$.  Then $\mathbb{R}^n - W$ is homotopy equivalent to the sphere $S^{n-k-1}$.
To see what is going on, consider the case where $W$ is a line in $\mathbb{R}^3$.  First you can deformation retract $\mathbb{R}^3$ along the direction of the line to obtain a punctured plane, and then you can deformation retract the punctured plane to a circle.  This works in general: collapse $\mathbb{R}^n$ down to $\mathbb{R}^{n-k} - \{0\}$ and then deformation retract to a sphere.
This will directly handle your first problem (where you remove a two dimensional subspace from $\mathbb{R}^4$) and the space $A$ in the second (where you remove a zero dimensional subspace from $\mathbb{R}^4$).  
To handle the set $B$, notice that the set $xy = 0$ is equivalent to $x = 0$ or $y = 0$, so you are removing two transversely intersecting planes from $\mathbb{R}^4$.  It may help to first work out what's going on when you remove two transversely intersecting lines from $\mathbb{R}^3$; you can deformation retract $B$ onto a familiar space, or you can use Mayer-Vietoris.
A: For (1), the homomorphism theorem isn't really useful here. It states that $\operatorname{Im} f$ is isomorphic to the quotient space $\mathbb{C}^2 / f^{-1}(0)$ as a vector space. And you are dealing with the set difference $\mathbb{C}^2 \setminus f^{-1}(0)$, not the quotient space.
To solve the problem, you can use two ideas. First, solve it in the special case when $f(x, y) = x$ is the coordinate form. In this case $\mathbb{C}^2 \setminus f^{-1}(0) = (\mathbb{C} \setminus \{0\}) \times \mathbb{C}$. Now $\mathbb{C} \setminus \{0\}$ is homotopy equivalent to $S^1$, and $\mathbb{C}$ is equivalent to a point, hence the product is equivalent to $S^1$.
The second idea is that the general case, when $f$ is not as simple as that, is in fact no different from the special case above. All you need is a good coordinate substitution.
For (2). My suggestion is to prove that $A$ is homotopy equivalent to $S^3$, and $B$ is homotopy equivalent to the torus $T^2$.
