Question

Which of the following statements is/are true?
(I) The function \(f(z)=\sin\left(\frac{1}{\cos(1/z)}\right)\) has an isolated singularity at \(z=0\).
(II) The value of the integral \(\oint_{|z|=1}\frac{dz}{z\sin(z)}\) is zero.

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

Check if the singular points of sin(1/cos(1/z)) cluster at 0, then expand 1/(z sin z) as a Laurent series.

Answer 1

The Correct Option is B

Step 1: Locate the points where the inner function blows up.
Write \(w=1/z\). The function \(\cos(1/z)=\cos w\) vanishes whenever \(w=\frac{\pi}{2}+k\pi\) for integer \(k\), that is at \(z_k=\frac{1}{\frac{\pi}{2}+k\pi}=\frac{2}{(2k+1)\pi}\).

Step 2: Check whether these points accumulate at z = 0.
As \(k\to\infty\), \(z_k\to 0\), so infinitely many points \(z_k\) lie in every punctured disk around \(z=0\). At each \(z_k\), \(\cos(1/z)\) has a simple zero, so \(1/\cos(1/z)\) has a pole there, and \(\sin\big(1/\cos(1/z)\big)\) has an essential singularity at each \(z_k\).

Step 3: Conclude on statement (I).
Since the singular points \(z_k\) cluster at \(z=0\), no punctured neighborhood of \(z=0\) is free of other singularities. Hence \(z=0\) is a non-isolated singularity, not an isolated one. Statement (I) is false.

Step 4: Expand \(1/(z\sin z)\) for statement (II).
Only \(z=0\) lies inside \(|z|=1\) (the next zero of \(\sin z\) is at \(z=\pi>1\)). Near \(z=0\), \[\sin z = z-\frac{z^3}{6}+\frac{z^5}{120}-\cdots \implies \frac{\sin z}{z}=1-\frac{z^2}{6}+\frac{z^4}{120}-\cdots\] so \[\frac{1}{z\sin z}=\frac{1}{z^2}\cdot\frac{1}{1-\frac{z^2}{6}+\cdots}=\frac{1}{z^2}+\frac{1}{6}+\frac{7z^2}{360}+\cdots\]

Step 5: Read off the residue.
The function \(1/(z\sin z)\) is even (only even powers of \(z\) appear), so the coefficient of \(z^{-1}\) is \(0\). By the residue theorem, \[\oint_{|z|=1}\frac{dz}{z\sin z}=2\pi i\cdot(\text{Res}_{z=0})=2\pi i\cdot 0=0.\] Statement (II) is true.

Conclusion: (I) is false and (II) is true, so only (II) holds. \[\boxed{\text{Only (II)}}\]