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Evaluate $$\lim_{R\to \infty}\int_{\gamma(0;R)}\frac{p(z)}{q(z)}\mathop{dz}$$ where $p$ and $q$ are polynomials and $\mathrm{deg}\ p<\mathrm{deg}\ q-1$.

Well what I tried to do was find an estimate for the quotient $p(z)/q(z)$. Let $$p(z)=\sum_{n=0}^m a_nz^n,\; q(z)=\sum_{n=0}^kb_kz^k$$ where $m<k-1$. Then, letting $\gamma(0;R)=Re^{it}\quad (0\leq t \leq 2\pi)$ $$\left \vert \frac{p(z)}{q(z)}\right \vert=\left \vert\frac{\sum_{n=0}^ma_nz^n}{\sum_{n=0}^kb_nz^n}\right \vert \leq \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_0|-|b_1|R-\dots - |b_k|R^k}$$ Now, by the estimation theorem: $$\left \vert\int_{\gamma(0;R)}\frac{p(z)}{q(z)}\mathop{dz}\right \vert \leq \int_0^{2\pi}R \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_0|-|b_1|R-\dots - |b_k|R^k}\mathop{dt}$$ and since $m+1<k$, then the integrand tends to $0$ as $R\to \infty$ so $$\lim_{R\to \infty}\int_{\gamma(0;R)}\frac{p(z)}{q(z)}\mathop{dz}=0$$

But I'm not sure if this is quite right or if I let some parts unjustified. Thanks in advance!

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This is almost correct, but be careful with the denominator of your fraction: $$\left \vert\frac{\sum_{n=0}^ma_nz^n}{\sum_{n=0}^kb_nz^n}\right \vert \leq \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_0|-|b_1|R-\dots - |b_k|R^k}$$ will actually be false for large $R$ because the denominator of the right-hand side will be negative (since the dominating term will be $|b_k|R^k$). What you can say is that if $R$ is large enough so that $|b_k|R^k>|b_0|+\dots+|b_{k-1}|R^{k-1}$, then by the reverse triangle inequality $$\left|\sum b_n z^n\right|\geq |b_k|R^k-|b_{k-1}|R^{k-1}-\dots-|b_0|>0$$ and so $$\left \vert\frac{\sum_{n=0}^ma_nz^n}{\sum_{n=0}^kb_nz^n}\right \vert \leq \frac{|a_0|+|a_1|R+\dots + |a_m|R^m}{|b_k|R^k-|b_{k-1}|R^{k-1}-\dots-|b_0|}.$$ Note that it is essential that we know that $|b_k|R^k-|b_{k-1}|R^{k-1}-\dots-|b_0|>0$, since otherwise the direction of the inequality might flip when we divide.

You might also want to justify more why the integrand goes to $0$ as $R\to\infty$. You can explain this by proving, for instance, that for $R$ sufficiently large, the numerator will be at most $2|a_m|R^m$ and the denominator will be at least $\frac{1}{2}|b_k|R^k$.

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  • $\begingroup$ Thanks you! I'll make sure to be more careful using the reverse triangle inequality in the future. Just one thing. To show that the integrand goes to zero, I've seen some exercises that use the $O$ notation but don't know how to use it properly. Could I say that the integrand is at most $O(R^{-1})$ to justify the limit? $\endgroup$ – user313212 Aug 27 '16 at 16:41
  • $\begingroup$ You could say that, but saying that in itself is not a proof--you would need to prove that it's at most $O(R^{-1})$ (though depending on context, this might be considered "obvious" enough that you don't need to write out the details). $\endgroup$ – Eric Wofsey Aug 27 '16 at 16:47

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