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Monday, June 18, 2018

About the singularities of $\sin \frac{1}{\cos\frac{1}{z}}$


Exercise. Find and classify the singularities of $f(z)=\sin\frac{1}{\cos\frac{1}{z}}$.
Solution. Whenever $\cos \frac{1}{z}$ is holomorphic and doesn't vanish $f(z)$ is holomorphic. Since all the zeros of $\cos z$ are its real zeros, i.e. $z=(2k+1)\frac{\pi}{2}$, then the only singularities of $f(z)$ are when  $z=0$, and when $z=\frac{1}{(2k+1)\frac{\pi}{2}}$. It may happen that one or more of them are removable, but this is not the case.

Let's first consider $z_0=0$. In this case, the singularity is not isolated.

Now consider $z_0=(2k+1)\frac{\pi}{2}$ for some $k$. Then if $z\to z_0$ along the real axis, $\cos \frac{1}{z}\to 0$ along the real axis, and if $z\to z_0$ in such a way that $\frac{1}{z}\to \frac{1}{z_0}$ along the line $\frac{1}{z_0}+ti$, then $\cos\frac{1}{z}\to 0$ along the imaginary axis. This, because $$\cos \frac{1}{z}=\cos\left(\frac{1}{z_0}+ti\right)=\cos\left(\frac{1}{z_0}\right)\cosh(t)-i\sin\left(\frac{1}{z_0}\right)\sinh(t)=\pm i\sinh(t)$$ where the sign depends on $k$.

So, in these cases, $\frac{1}{\cos\frac{1}{z}}\to \infty$ along the real axis and the imaginary axis respectively. Along the real axis, $f(z)$ oscilates, and in particular, is bounded, but along the imaginary axis, $f(z)\to\infty$. So these singularities are essential.
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