Step 1: For (A), use the general fact that \(\operatorname{Arg}(zw)\equiv\operatorname{Arg}(z)+\operatorname{Arg}(w)\pmod{2\pi}\), but the principal argument is restricted to \((-\pi,\pi]\). Whenever \(\operatorname{Arg}(z)+\operatorname{Arg}(w)\) falls outside \((-\pi,\pi]\), the principal value wraps around by \(\pm2\pi\), so \(\operatorname{Log}(zw)\) differs from \(\operatorname{Log}(z)+\operatorname{Log}(w)\) by \(2\pi i\). Since this wrap-around genuinely occurs for some \(z,w\), the identity cannot hold for all \(z,w\), so (A) is false.
Step 2: For (B), solve using the periodicity of the complex exponential instead of splitting into real and imaginary parts. Since \(\cos\pi+i\sin\pi=-1\), \(z_0=\pi\) is one solution of \(e^{iz}=-1\). Because \(e^{iz}=e^{iz_0}\) if and only if \(i(z-z_0)=2k\pi i\) for some \(k\in\mathbb{Z}\), i.e. \(z=z_0+2k\pi\), the full solution set is \(z=\pi+2k\pi=(2k+1)\pi\), \(k\in\mathbb{Z}\), matching option (B).
Step 3: For (C), use the modulus formula \(|\cos z|^2=\cos^2x\cosh^2y+\sin^2x\sinh^2y\). Since \(\sinh^2y=\cosh^2y-1\), \[|\cos z|^2=\cos^2x\cosh^2y+\sin^2x(\cosh^2y-1)=\cosh^2y-\sin^2x.\] As \(y\to\infty\), \(\cosh^2y\to\infty\) while \(\sin^2x\) stays bounded between 0 and 1, so \(|\cos z|\to\infty\). This confirms \(\cos(z)\) is unbounded, so (C) is false.
Step 4: For (D), write a general Mobius map as \(w=(az+b)/(cz+d)\) with \(ad-bc\neq0\). Fixed points satisfy \(z(cz+d)=az+b\), i.e. \[cz^2+(d-a)z-b=0.\] If \(c\neq0\) this is a genuine quadratic with at most 2 roots; if \(c=0\) it is linear (or identically zero) with at most 1 root, unless \(d=a\) and \(b=0\) as well, in which case every \(z\) is fixed and the map is the identity \(w=z\). So three or more fixed points force the identity map, not a constant, confirming (D) is false.
Step 5: Only (B) survives all checks. \[\boxed{\text{(B)}}\]
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: