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.

Question

Match List-I with List-II and choose the correct option:

LIST-I (Function)	LIST-II (Value)
(A) $ \int_{\gamma} \frac{1}{z-a} \, dz $, where $ \gamma: \|z-a\|=r, r > 0 $	(III) $ 2i\pi $
(B) $ \int_{\gamma} \frac{z+2}{z} \, dz $, where $ \gamma: z = 2e^{it}, 0 \le t \le \pi $	(IV) $ i\pi $
(C) $ \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz $, where $ \gamma: \|z\|=3 $	(II) $ 2i\pi(e^4 - e^2) $
(D) $ \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz $, where $ \gamma: \|z\|=2 $	(I) $ -4 + 2i\pi $

Choose the correct answer from the options given below:

Answer 1

Alternate approach.

For statement (I), substitute $w=1/z$ so the behaviour near $z=0$ becomes the behaviour of $g(w)=\sin(1/\cos w)$ as $w\to\infty$. The function $\cos w$ has infinitely many zeros $w=\frac{\pi}{2}+k\pi$, $k\in\mathbb{Z}$, spread across the whole $w$-plane, and each gives a point where $g$ has an essential singularity. Translating back through $z=1/w$, these correspond to points $z_k=\frac{2}{(2k+1)\pi}$ that pile up at $z=0$ as $|k|\to\infty$. A singularity is isolated only if some punctured disk around it contains no other singular point; here every punctured disk about $0$, however small, contains infinitely many $z_k$. So $z=0$ fails the isolation test and statement (I) is false.

For statement (II), use the standard Laurent (cosecant) expansion valid for $0<|z|<\pi$: \[\csc z=\frac{1}{z}+\frac{z}{6}+\frac{7z^3}{360}+\cdots\] Then \[\frac{1}{z\sin z}=\frac{1}{z}\csc z=\frac{1}{z}\left(\frac{1}{z}+\frac{z}{6}+\frac{7z^3}{360}+\cdots\right)=\frac{1}{z^2}+\frac{1}{6}+\frac{7z^2}{360}+\cdots\] There is no $z^{-1}$ term, so the residue at the only enclosed singularity $z=0$ is $0$, and by Cauchy's residue theorem \[\oint_{|z|=1}\frac{dz}{z\sin z}=2\pi i(0)=0.\] This confirms statement (II) is true while (I) is false, so the correct choice is only (II). \[\boxed{\text{Only (II)}}\]

LIST-I (Function)	LIST-II (Value)
(A) \( \int_{\gamma} \frac{1}{z-a} \, dz \), where \( \gamma: \|z-a\|=r, r > 0 \)	(III) \( 2i\pi \)
(B) \( \int_{\gamma} \frac{z+2}{z} \, dz \), where \( \gamma: z = 2e^{it}, 0 \le t \le \pi \)	(IV) \( i\pi \)
(C) \( \int_{\gamma} \frac{e^{2z}}{(z-1)(z-2)} \, dz \), where \( \gamma: \|z\|=3 \)	(II) \( 2i\pi(e^4 - e^2) \)
(D) \( \int_{\gamma} \frac{z^2 - z + 1}{2(z-1)} \, dz \), where \( \gamma: \|z\|=2 \)	(I) \( -4 + 2i\pi \)

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.

Show Hint

The Correct Option is B

Solution and Explanation

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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.