I've been asked to prove the following;
$$\lim_{R \rightarrow \infty}\int_{C_R} \frac{z^2 + 8z + 42}{(z^2+4)(z^2-4z+5)}dz=0$$
Given that $C_R$ is a circle of radius $R$ centered at $0$.
I thought the easiest way to prove this would be using the fact that;
$$|\int_C f(z)dz| \le ML$$
Where $M$ is the upper boundary of the integral, and $L$ is the arc length.
Firstly, I noted; $$|\frac{z^2 + 8z + 42}{(z^2+4)(z^2-4z+5)}| \le \frac{|z^2 + 8z + 42|}{|(z^2+4)(z^2-4z+5)|}$$ And then that; $$|z^2 + 8z + 42| \le |z|^2 + 8|z| +42 = |R^2 e^{2i\theta}| + 8|Re^{i\theta}|+42 = R^2 + 8R + 42$$ $$|(z^2+4)(z^2-4z+5)| = |z^4 -4z^3 +9z^2-16z+20| \ge||z|^4 -4|z|^3 +9|z|^2-16|z|+20|$$ $$=|R^4 -4R^3 +9R^2-16R+20| = R^4 -4R^3 +9R^2-16R+20$$
Then, I have the following; $$|\frac{z^2 + 8z + 42}{(z^2+4)(z^2-4z+5)}| \le \frac{R^2 + 8R + 42}{R^4 -4R^3 +9R^2-16R+20} = M$$
Now, it's fairly obvious that the limit of $M$ will go to zero as $R$ approaches $\infty$, so is it appropriate to conclude the following?? $$\lim_{R \rightarrow \infty} |\int_{C_R} \frac{z^2 + 8z + 42}{(z^2+4)(z^2-4z+5)}dz| \le \lim_{R \rightarrow \infty} ML = 0$$ Thus, is it appropriate to say that as the maximum value the function can take is zero, that the entire integral is equal to zero??
Further, we've been asked to use this to show that the same integral over the contour $C$, where $C$ is a circle of radius 5, centre 2, is also equal to zero. Is it appropriate to say that what I've determined will hold for a circle of any given radius, regardless of the point at which it is centered, so that the statement is true??
Any help would be fantastic!!