I was trying out this problem from Tom Dieck's book on Algebraic Topology. It says if $f_1, \dots, f_k : \mathbb{C}^n \to \mathbb{C}$ are linearly independent linear functionals $(1 \leq k \leq n)$. Then the complement $ \mathbb{C}^n \setminus \bigcup_{j=1}^k \ker f_j $ is homotopic equivalent to $\underbrace{S^1 \times \cdots \times S^1}_{k \mbox{ times}}$.
We have $a_1, \dots, a_k \in \mathbb{C}^n$ such that $f_j(z) = a_j^t z$ for all $z \in \mathbb{C}^n$ and $j = 1, \dots, k$. Now the obvious map that I can think of is the following $$ \varphi : \mathbb{C}^n \setminus \bigcup_{j=1}^k \ker f_j \to \prod_{j=1}^k S^1 $$ taking $z \mapsto \left( \frac{a_1^t z}{\| a_1^j z \|}, \cdots, \frac{a_k^tz}{\|a_k^tz\|} \right)$. But don't know whether this map gives us a homotopic equivalence. The other observation that I have in mind is this, since we are given $f_j$'s are linearly independent it tells us that $\dim_{\mathbb{C}}\left(\bigcap_{j=1}^k \ker f_j\right) = n-k$, though I don't whether this is at all needed. Any help would be greatly appreciated.