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I'm having trouble with the following complex integral:

$$\int_{C}^{}\frac{dz}{\sqrt{1 - z^{2}}}$$ where C is a positively oriented ellipse $${x^{2}\over a^{2}} +{ y^{2}\over b^{2}} = 1$$ where $$a^{2} - b^{2} = 1$$

I know z(t) = $a\cos(t) + ib\sin(t)$ but I'm not sure about how to factor after plugging z into the integrand and simplifying using the fact that $$a^{2} - b^{2} = 1$$

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up vote 4 down vote accepted

As you say: $z(t) = a\cos t + ib\sin t$ gives a bijection from $[0,2\pi)$ to the ellipse. Given the definition of $z$, we can see that $\operatorname{d}\!z=(-a\sin t + ib\cos t) \, \operatorname{d}\!t$. Hence:

$$\oint_C \frac{\operatorname{d}\!z}{\sqrt{1-z^2}} = \int_0^{2\pi} \frac{-a\sin t + ib\cos t }{\sqrt{1-(a\cos t + ib\sin t)^2}}\, \operatorname{d}\!t$$

The trick here is to use the fact that $a^2-b^2=1$, or rather $1=a^2-b^2$. Notice that

\begin{array}{ccc} 1-(a\cos t + ib\sin t)^2 &\equiv& 1- a^2\cos^2t +b^2\sin^2t - 2iab\cos t \sin t \\ \\ &\equiv& a^2-b^2 - a^2\cos^2t +b^2\sin^2t - 2iab\cos t \sin t \\ \\ &\equiv& a^2(1-\cos^2t) -b^2(1-\sin^2) - 2iab\cos t \sin t \\ \\ &\equiv& a^2\sin^2t - b^2\cos^2t - 2iab\cos t \sin t \\ \\ &\equiv& (-a\sin t + ib\cos t)^2 \end{array}

It follows that:

$$\oint_C \frac{\operatorname{d}\!z}{\sqrt{1-z^2}} = \int_0^{2\pi} \frac{-a\sin t + ib\cos t }{\sqrt{(-a\sin t + ib\cos t)^2}}\, \operatorname{d}\!t$$

This integrand has an explicit anti-derivative:

\begin{array}{ccc} \int_0^{2\pi} \frac{-a\sin t + ib\cos t }{\sqrt{(-a\sin t + ib\cos t)^2}}\, \operatorname{d}\!t &=& \left[ t \, \frac{-a\sin t + ib\cos t}{\sqrt{(-a\sin t + ib\cos t)^2}} \right]_0^{2\pi} \\ \\ &=& 2\pi i \frac{b}{\sqrt{-b^2}} \\ \\ &=& 2\pi \frac{b}{|b|} \end{array}

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Good job. (+1).. – Ron Gordon Apr 21 '13 at 22:17

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