# Problem with residues in Complex Analysis?

While dealing with residues in Complex Analysis, many books don't care to define the function on singular points. I think it is not rigorous enough. Without defining the function on a point, it is unfair to do integration. In many cases, emphasis on why the point is a singularity gets lost. For example

Complex Variables and Applications by Brown & Churchill 8e Section 69 Example 1 finds

$$\int_C z^2\sin{\left(\frac1z\right)}dz$$.

It directly says $$z=0$$ is singular point.

If I define $$f(0)=z_0\neq0$$, then $$z=0$$ is a singularity because $$f$$ is not continuous at $$z=0$$. But if I define $$f(0)=0$$, then $$f$$ is continuous but $$f^\prime$$ is not.

• You seem to think that $\sin (\frac 1 z)$ is bounded. It is not. Commented Nov 23, 2023 at 12:26
• But $\lim_{z\to0}z^2\sin{\frac1z}=0$ Commented Nov 23, 2023 at 12:34
• Your statement is only true if you take the limit over real $z$. Don't forget you're working with functions of complex variables. Commented Nov 23, 2023 at 13:12
• @RameshwarSao If we have a removable singularity then it is always assumed that we define $f$ on this point to be the limit at the point, which makes $f$ a holomorphic function. But in your example, the point $z=0$ is an essential singularity, the worst possible case. There is no way to define $f(0)$ in this case to make it continuous. As mentioned above, $\lim\limits_{z\to 0} z^2\sin(\frac{1}{z})$ is not $0$. Take for example the sequence $z_n=\frac{1}{in}$ and check.
– Mark
Commented Nov 23, 2023 at 14:15
• What is $C$? An integral like this is best evaluated by taking $w=1/z$. If for example $C$ is $|z|=r$ anticlockwise for some $r>0$, the new contour would be $|z|=1/r$ clockwise.
– J.G.
Commented Nov 25, 2023 at 0:02