For the linear vector field $f(x,y) = (f_1, f_2) = (ax+by,cx+dy)$, show that the index with respect to the origin is $\pm 1$ depending on whether $ad-bc > 0$ or $ad-bc < 0$.
I've gone through this calculation four or five times now and I get a different answer every time. Please help me figure out what I'm doing wrong! Here's what I've tried:
By definition, the index of a critical point is defined as the index of any Jordan curve that surrounds the point. (I assume the critical point is isolated and unique.) To find the index of a loop $\gamma$, I use $$Ind(\gamma,f) = \frac{1}{2 \pi} \int_\gamma \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} $$ and I choose my Jordan curve $\gamma$ to be the square with corners in $(1,1), (1,-1),(-1,1)$ and $(-1,-1)$. Then I can split $\gamma$ into four pieces: \begin{align*} & \gamma_1 = \{(x,-1): -1 \leq x \leq 1\} \\ & \gamma_2 = \{(1,y): -1 \leq y \leq 1\} \\ & \gamma_3 = \{(x,1): -1 \leq x \leq 1\} \\ & \gamma_4 = \{(y,-1): -1 \leq x \leq 1\} \end{align*}
The next step is to integrate along each of these line segments: \begin{align*} & \int_{\gamma_1} \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} = - \arctan \left( \frac{c-d}{a-b} \right) + \arctan \left( \frac{-c-d}{-a-b} \right) \\ & \int_{\gamma_2} \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} = \arctan \left( \frac{c+d}{a+b} \right) - \arctan \left( \frac{c-d}{a-b} \right) \\ & \int_{\gamma_3} \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} = \arctan \left( \frac{c+d}{a+b} \right) - \arctan \left( \frac{-c+d}{-a+b} \right) \\ & \int_{\gamma_4} \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} = \arctan \left( \frac{-c+d}{-a+b} \right) - \arctan \left( \frac{-c-d}{-a-b} \right) \\ \end{align*} Now, adding these four integrals together I get $$Index(\gamma,f) = \frac{1}{\pi} \left( \arctan \left( \frac{c+d}{a+b} \right)- \arctan \left( \frac{c-d}{a-b} \right) \right) $$ and finally I use the addition formula for $\arctan$ to get
$$Index(\gamma,f) = \frac{1}{\pi}\arctan \left( \frac{2(ad-bc)}{a^2-b^2+c^2-d^2} \right).$$
Here I can see that the sign of $ad-bc$ determines the sign of the index, but the expression doesn't seem integer-valued at all to me. What's wrong?
EDIT: Here's how I solved the integrals along the $\gamma_i$. Use that $df_1 = a\cdot dx = b \cdot dy$ and $df_2 = c \cdot dx = d \cdot dy$. On $\gamma_1$ we have $y = -1$, so we integrate with respect to $x$ and get \begin{align*} \int_{\gamma_1} \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} =\int_{\gamma_i} \frac{(ax+by)c\cdot dx-(cx+dy)a\cdot dx}{(ax+by)^2+(cx+dy)^2} = \int_{-1}^1 \frac{(bc-ad)y}{(ax+by)^2+(cx+dy)^2}dx = \int_{-1}^1 \frac{\frac{(bc-ad)y}{(ax+by)^2}}{1+(\frac{cx+dy}{ax+by})^2}dx \end{align*} But the derivative of the arctangent is $$\frac{d}{dx}\left( \arctan\left(\frac{cx+dy}{ax+by}\right) \right) = \frac{1}{1+\left( \frac{cx+dy}{ax+by}\right)^2}\cdot\frac{(bc-ad)y}{(ax+by)^2}$$ Hence we must have $$\int_{\gamma_1} \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} = \left[ \arctan \left( \frac{cx-d}{ax-b}\right) \right]_{x=-1}^1=\arctan \left( \frac{c-d}{a-b}\right) - \arctan \left( \frac{-c-d}{-a-b}\right)$$ Actually, it looks like I got the signs reversed earlier, where I claimed that $$\int_{\gamma_1} \frac{f_1 df_2 - f_2 df_1}{f_1^2 + f_2^2} = - \arctan \left( \frac{c-d}{a-b} \right) + \arctan \left( \frac{-c-d}{-a-b} \right)$$ but if I change the signs then all terms cancel and I get $0$, which is still not the right answer. I'm very confused now!