Consider the following statements:
(I) Residue of $f(z)=e^z$ at $z=\infty$ is $-1$.
(II) If a meromorphic function $f$ has 5 simple zeros and 2 simple poles inside $|z|=1$, then $\oint_{|z|=1}\frac{f'(z)}{f(z)}dz = 6i\pi$.
Pick out the correct option.

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), use the direct contour definition of the residue at infinity: for any function analytic in a punctured neighborhood of infinity, \[\text{Res}_{z=\infty} f = -\frac{1}{2\pi i}\oint_{|z|=R} f(z)\,dz\] for $R$ large enough to enclose every finite singularity of $f$, traversed counterclockwise. Since $f(z)=e^z$ is entire, it has no finite singularities at all, so Cauchy's theorem gives $\oint_{|z|=R} e^z\,dz=0$ for every $R$. Hence \[\text{Res}_{z=\infty}e^z=-\frac{1}{2\pi i}(0)=0,\] which contradicts the claimed value $-1$, so statement (I) is false.

For statement (II), build the result from local behaviour instead of quoting the argument-principle formula outright. Near a simple zero $a_i$ of $f$, write $f(z)=(z-a_i)g(z)$ with $g(a_i)\ne 0$; then \[\frac{f'(z)}{f(z)}=\frac{1}{z-a_i}+\frac{g'(z)}{g(z)},\] so $f'/f$ has a simple pole at $a_i$ with residue $1$. Near a simple pole $b_j$, write $f(z)=\frac{h(z)}{z-b_j}$ with $h(b_j)\ne 0$; then \[\frac{f'(z)}{f(z)}=\frac{-1}{z-b_j}+\frac{h'(z)}{h(z)},\] so $f'/f$ has residue $-1$ at $b_j$. Summing over the 5 zeros and 2 poles inside $|z|=1$ gives total residue $5(1)+2(-1)=3$. By the residue theorem, \[\oint_{|z|=1}\frac{f'(z)}{f(z)}dz=2\pi i(3)=6\pi i=6i\pi.\] So statement (II) is true, confirming (I) false and (II) true. \[\boxed{\text{(I) is false, but (II) is true}}\]

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) \( -4 + 2i\pi \)

Consider the following statements:
(I) Residue of \(f(z)=e^z\) at \(z=\infty\) is \(-1\).
(II) If a meromorphic function \(f\) has 5 simple zeros and 2 simple poles inside \(|z|=1\), then \(\oint_{|z|=1}\frac{f'(z)}{f(z)}dz = 6i\pi\).
Pick out the correct option.

Show Hint

The Correct Option is B

Solution and Explanation

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Consider the following statements:(I) Residue of \(f(z)=e^z\) at \(z=\infty\) is \(-1\).(II) If a meromorphic function \(f\) has 5 simple zeros and 2 simple poles inside \(|z|=1\), then \(\oint_{|z|=1}\frac{f'(z)}{f(z)}dz = 6i\pi\).Pick out the correct option.