I'm working on the following problem:

Given $a,b>0$, define the path $\gamma$ whose image is an ellipse. $$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$$ traced counterclockwise. By showing that $\int_{\gamma}z^{-1}dz = \int_{\gamma}z^{-1}dz$ for a suitable circle show that $$\int_0^{2\pi}\frac{1}{a^2\cos^2{t}+b^2\sin^2{t}}dt=\frac{2\pi}{ab}$$

Attempt 1: Suppose $\gamma(t)= a \cos(t) + ib\sin(t)$ then we obtain $$\int_0^{2\pi}\frac{-a \sin(t) + ib\cos(t)}{a \cos(t) + ib\sin(t)}dt$$ I tired rationalizing the numerator but end up with $a^2+b^2$ in the numerator and a mess in the denominator. So I'm unable to obtain the LHS.

Attempt 2: Recognize $$\int_{\gamma}\frac{1}{z}=\int\frac{\gamma'(t)}{\gamma(t)}dt=\int\frac{\partial}{\partial t}\log(\gamma(t))dt$$ so that $$\frac{\partial}{\partial t}\log(\gamma(t)) = \frac{1}{a^2\cos^2{t}+b^2\sin^2{t}}$$ Integrating in $t$ and exponentiating to solve for $\gamma$ gives a mess. It doesn't reduce either. I think here, there also maybe an issue of branch cuts that I haven't considered carefully.

I also recognize that the denominator factors: $(a \cos(t) + ib\sin(t))(a \cos(t) - ib\sin(t))$ I haven't been able to use this information though.

Any ideas? Thanks!


2 Answers 2


You should rationalize the denominator, not the numerator. You have \begin{align} \int_0^{2\pi}\frac{-a \sin t + ib\cos t }{a \cos t + ib\sin t }dt &=\int_0^{2\pi}\frac{(-a \sin t + ib\cos t )(a\cos t-ib\sin t)}{a^2\cos^2t+b^2\sin^2t}dt\\ \ \\ &=\int_0^{2\pi}\frac{(-a^2+b^2)\sin t\cos t+iab}{a^2\cos^2t+b^2\sin^2t}dt\\ \ \\ \end{align} Now is the time to use the suggestion: this integral will be equal to $\int_{\gamma'}\frac1z\,dz$ if $\gamma'$ is small enough and surrounds the origin. So take $\gamma'$ to be a small circle of radius $r$, and calculate (easily) that $\int_{\gamma'}\frac1z\,dz=2\pi i$.

As both integrals should be equal (because combining both you enclose a region where $1/z$ is analytic), you get $$ \int_0^{2\pi}\frac{(-a^2+b^2)\sin t\cos t+iab}{a^2\cos^2t+b^2\sin^2t}dt=2\pi i. $$ This tells you that the real part on the left-hand-side is zero. Comparing the imaginary parts, you get $$ \int_0^{2\pi}\frac{ab}{a^2\cos^2t+b^2\sin^2t}dt=2\pi. $$

  • 1
    $\begingroup$ How did you know that the resulting integral was equal to $\int_{\gamma'}\frac{1}{z}dz$? Is that how I was supposed to interpret the given hint? Also why do you say "small enough"? There's only one pole for $1/z$, can't the contour be of any size so long as its closed? $\endgroup$
    – yoshi
    Feb 11, 2018 at 1:50
  • 1
    $\begingroup$ I say "small enough" because it needs to fit entirely inside the ellipse; it could also be "big enough" so that it contains the ellipse. If the ellipse and the circle do not intersect, they form a closed region where $1/z$ is analytic, so the integral over the whole boundary (which is the integral over one of the two curves minus the integral over the other) is zero. $\endgroup$ Feb 11, 2018 at 1:54

Here is a bit more "Complex Analysis" solution. Consider the simple graph:

enter image description here

Here we have our ellipse $\gamma$, small enough circle $\delta$, and two segments $I_1$ and $I_2$.

$\gamma$ is homotopically equivalent to the contour $I_1 \rightarrow \delta \rightarrow I_2$.


$$ \int_{\gamma} dz/z = \int_{I_1} dz/z + \int_{\delta} dz/z + \int_{I_2} dz/z $$

Integrals over $I_1$ and $I_2$ cancels out. Now it remains to take the integral $dz/z$ over a circle $\delta$.


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