Given the function $f$ with the rule $f(z) = \frac{z \sinh\left(\frac{1}{z^2}\right)}{z^2 + 1}$.
(a) Determine and classify the singular points of the function $f$ and calculate the residues at these points.
(b) Given a real number $r \in (0, \infty) \setminus \{1, 2\}$, compute the integral $\int_K f(z) \, dz$, where $K = \{z \in \mathbb{C} \mid |z + i| = r\}$ is a positively oriented circle.
Attempt: (a) Singular Points and Residues of $f(z)$
To determine and classify the singular points of the function
$$ f(z) = \frac{z \sinh\left(\frac{1}{z^2}\right)}{z^2 + 1}, $$
I need to analyze where the function might become undefined. The singularities occur where the denominator $z^2 + 1$ is zero or where the argument of the $\sinh$ function has a singularity.
-
Denominator Singularities:
The equation $z^2 + 1 = 0$ has solutions at $z = \pm i$, which are simple poles since the denominator linearly vanishes at these points. -
Singularity at $z = 0$:
The function $\sinh\left(\frac{1}{z^2}\right)$ has an essential singularity at $z = 0$ due to the $\frac{1}{z^2}$ term inside the hyperbolic sine. Therefore, $z = 0$ is an essential singularity for $f(z)$. -
Residue at $z = \pm i$:
To find the residues at $z = i$ and $z = -i$, we compute:
$$
\text{Residue at } z = i: \quad \text{Res}(f, i) = \lim_{z \to i} (z – i) f(z) = \lim_{z \to i} \frac{z \sinh\left(\frac{1}{z^2}\right)}{z + i} = \frac{i \sinh\left(\frac{1}{-1}\right)}{2i} = \frac{\sinh(-1)}{2}
$$
$$
\text{Residue at } z = -i: \quad \text{Res}(f, -i) = \lim_{z \to -i} (z + i) f(z) = \lim_{z \to -i} \frac{z \sinh\left(\frac{1}{z^2}\right)}{z – i} = \frac{-i \sinh\left(\frac{1}{-1}\right)}{-2i} = \frac{\sinh(-1)}{2}
$$
Since $\sinh(-1) = -\sinh(1)$, the residues are:
$$
\text{Res}(f, i) = \frac{-\sinh(1)}{2}, \quad \text{Res}(f, -i) = \frac{-\sinh(1)}{2}
$$
(b) Integral Computation on Circle $K$
How can I even do it and also have I correctly solved (a) part of the exercise?
Best Answer
OK so now you actually have finished most part of (a). What you haven't noted is that there is another possible residue: $z=0$ for you cannot have $\sinh\frac 10$!
So let's complete (a) first:
If we want to get the residue of $f$ at $0$, we just need to expand $f$ at $0$ and see what its coefficients of the term $\frac 1z$ looks like:
$\begin{aligned} \frac{z\sinh(\frac1{z^2})}{1+z^2}&=z\cdot(\frac1{z^2}+\frac{1}{3!\cdot z^6}+\frac{1}{5!\cdot z^{10}}+\cdots)(1-z^2+z^4-z^6+\cdots) \end{aligned}$
The term $\frac1z$ comes from such product:$z\cdot\frac{1}{z^2}\cdot1,z\cdot\frac{1}{3!z^6}\cdot z^4,z\cdot\frac{1}{5!z^{10}}\cdot z^{8},z\cdot\frac{1}{7!z^{14}}\cdot z^{12}\cdots$
Hence the coefficient of $\frac1z$ is:$1+\frac 1{3!}+\frac 1{5!}+\frac 1{7!}+\cdots=\sinh1$. (just the Taylor series of $\sinh x$ at $x=1$)
Now we can solve (b):
The residue theorem states that the intergral on a contour equals its summation of residues inside the contour and multiply $2\pi i$. So we just need to figure out whta the residues inside the contour looks like.
The above graph shows that if $r<1$, then the only residue inside the contour is just $z=-i$, in this situation we have $\int_K f(z)\mathrm dz=2\pi i\sum Res(f)=2\pi i\cdot Res(f,-i)=2\pi i\cdot \frac{-\sinh(1)}{2}= -\pi \sinh(1)\cdot i$
When $1<r<2$, this graph shows that the contour hasn't contained $z=i$, but it has already contained $z=0$. However, the residue at $z=0$ is $0$, so it makes no contribution to the integral. Then we have $\int_K f(z)\mathrm dz=2\pi i\sum Res(f)=2\pi i\cdot (Res(f,-i)+Res(f,0))=2\pi i\cdot( \frac{-\sinh(1)}{2}+\sinh1)= \pi \sinh(1)\cdot i$
When $r>2$, we can see the contour now has contained all our sigular points:$z=i,z=0,z=-i$. This way we just need to add a new term to our integral: $\begin{aligned} \int_K f(z)\mathrm dz&=2\pi i\sum Res(f)=2\pi i\cdot (Res(f,i)+Res(f,0)+Res(f,-i))\\&=2\pi i\cdot( \frac{\sinh(-1)}{2}+\sinh1+\frac{-\sinh1}{2})= 0 \end{aligned}$
(Note that $\sinh(-z)=\sinh(z)$)