# Limit of complex integral

Let $\partial C_R(0)$ be the circle centered at $0$ with radius $R>0$. Show that

$\displaystyle \lim\limits_{R\to\infty}\oint_{\partial C_R(0)}\frac{d z}{z^3+z+1}=0$

Proof. It is sufficient to show that the absolute value converges. Choose $\gamma : [0,2\pi]\to\mathbb C,\ t\mapsto R\exp(it)$, then

$\displaystyle\left|\int_0^{2\pi}\frac{Rie^{it}}{R^3e^{3it}+Re^{it}+1}dt\right|\leq \int_0^{2\pi}\left|\frac{Rie^{it}}{R^3e^{3it}+Re^{it}+1}dt\right|=\int_0^{2\pi}\frac{Rdt}{|R^3e^{3it}+Re^{it}+1|}\leq\int_0^{2\pi}\frac{dt}{|R^2e^{3it}+e^{it}|}$

Unfortunately I can't think of another comparison to further simplify the integral but I'm sure not much is needed anymore.

• You need $+1/R$ downstairs in the last expression. – zhw. May 25 '17 at 15:08
• Isn't $|z+1|=\sqrt{(x+1)^2+y^2}\geq\sqrt{x^2+y^2}= |z|$? – user424862 May 25 '17 at 16:46
• Try that with $x=-1,y=0.$ – zhw. May 25 '17 at 16:49
• Oh jesus, you're right! – user424862 May 25 '17 at 17:08

Note that

$$\int_0^{2\pi}\frac{Rdt}{|R^3e^{3it}+Re^{it}+1|}= \frac{1}{R^2}\int_0^{2\pi}\frac{dt}{|e^{3it}+e^{it}/R^2+1/R^3|}.$$

The last integrand above converges uniformly to $1,$ so in the limit we get $0\cdot 2\pi = 0.$

• In this case though I would need something like Lebesgue's dominated convergence theorem to interchange $\lim$ and $\int$ but I haven't seen one for complex-valued functions in our lecture yet. – user424862 May 25 '17 at 15:21
• @user424862 No, we have uniform convergence! Monsieur Lebesgue can be left undisturbed. – zhw. May 25 '17 at 15:28
• So uniform convergence implies that I can just pull the lim under the integral? That's sweet and I didn't know that. – user424862 May 25 '17 at 15:59

You can affirm the following inequality: $$\int_0^{2\pi}\frac{dt}{|R^2e^{3it}+e^{it}|}\leq\frac{2\pi}{R^2-1}$$

This can be seen by recalling that $$|z_1-z_2|$$ is the distance between the two complex numbers $z_1$ and $z_2$. Thus to maximize the integrand you must minimize this distance. Notice that $R^2e^{3it}$ and $e^{it}$ are two circles in the complex plane of radii $R^2$ and $1$ respectively. Thus the minimum distance between these two circles is $R^2-1$.

The bounding function is in $O(R^{-2})$ and tends to zero as $R$ tends to infinity.

• Although I see your point, I'm not really happy about arguing this way as it seems a little wishy-washy. Isn't there some factor we can cheat into the integral which gives the wanted inequality without a big description that needs to be added? – user424862 May 25 '17 at 15:17

The function $f(z)=\frac{1}{z^3+z+1}$ is a meromorphic function with three simple poles within the region $|z|\leq 2$. By the residue theorem it follows that for any $R>2$ the value of the integral $$\oint_{|z|=R}\frac{dz}{z^3+z+1}$$ is always the same, i.e. $2\pi i$ times the sum of the residues of $f(z)$ at its poles. On the other hand if $|z|\geq 2$ we have $\left|f(z)\right|\leq \frac{3}{|z|^3}$ (that is a loose inequality, but we do not really need to be more accurate), hence: $$\forall R>2,\qquad \left|\oint_{|z|=R}\frac{dz}{z^3+z+1}\right|\leq \oint_{|z|=R}\frac{3}{|z|^3}\,dz = \frac{6\pi}{R^2}$$ and it follows that: $$\forall R>2,\qquad \oint_{|z|=R}\frac{dz}{z^3+z+1} = 2\pi i \sum_{i=1}^{3} \text{Res}(f,\zeta_i)=\color{red}{0}.$$