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Let $ F: U \subset \mathbb R ^2 \to \mathbb R^2$ be a map with $ F(x,y) = (f(x,y) , g(x,y))$ ,satisfies $F(-q)=-F(q) \quad \forall q \in D \subset U$ where $D$ is a closed disk with center the origin and set $\partial D := c$. Assume that $F$ has no zeros in $\partial D$.

(a) Prove that the index $ n(F; D)$ is an odd integer.

(b) Prove that $F$ has at least one zero on the disk $D$.

Definition: $\displaystyle{ n(F;D) =\frac{1}{2 \pi} \int_c \theta _0}$ where $\displaystyle{ \theta_0 = \frac{g df - fdg}{f^2 + g^2}}$

One can easily check that the index $n(F;D)$ is the winding number of the curve $ F \circ c$ about the origin and so index is an integer.

edit: I didn't write all hypothesis. I am really sorry for that. I hope now is clear.

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    Now I see! I'll think about it.2012-06-20

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The proof that the winding number of $F\circ c$ is odd is given in $f:\mathbb{S}^1\rightarrow\mathbb{S}^1$ odd $\Rightarrow$ $\mathrm{deg}(f)$ odd (Borsuk-Ulam theorem). By the argument principle, $F$ has an odd number of zeros in $D$, counting multiplicities. Hence, it has at least one zero.