Question

Which one of the following options is true?
(A) If \(\operatorname{Log}(z)\) denotes the principal value of the logarithm, then \(\operatorname{Log}(zw)=\operatorname{Log}(z)+\operatorname{Log}(w)\) for all \(z,w\in\mathbb{C}\setminus\{0\}\).
(B) The solution set of the equation \(e^{iz}=-1\) is \(\{(2k+1)\pi:k\in\mathbb{Z}\}\).
(C) \(\cos(z)\) is bounded in the whole complex plane.
(D) A Mobius transformation with three fixed points is a constant.

• A. If \(\operatorname{Log}(z)\) denotes the principal value of the logarithm, then \(\operatorname{Log}(zw)=\operatorname{Log}(z)+\operatorname{Log}(w)\) for all \(z,w\in\mathbb{C}\setminus\{0\}\)
• B. The solution set of the equation \(e^{iz}=-1\) is \(\{(2k+1)\pi:k\in\mathbb{Z}\}\)
• C. \(\cos(z)\) is bounded in the whole complex plane
• D. A Mobius transformation with three fixed points is a constant

Hint:

Test each option with a specific case or the explicit fixed-point equation.

Answer 1

The Correct Option is B

Step 1: Check (A) with a counterexample. Take \(z=w=-1\). The principal value \(\operatorname{Log}(-1)=i\pi\), so \(\operatorname{Log}(z)+\operatorname{Log}(w)=2i\pi\). But \(zw=1\), and \(\operatorname{Log}(1)=0\neq2i\pi\). So the identity fails in general, and (A) is false.

Step 2: Check (B). Write \(z=x+iy\). Then \(e^{iz}=e^{i(x+iy)}=e^{-y}e^{ix}=e^{-y}(\cos x+i\sin x)\). Since \(|{-1}|=1\), we need \(e^{-y}=1\), so \(y=0\). Then we need \(e^{ix}=-1\), i.e. \(\cos x=-1\) and \(\sin x=0\), which gives \(x=(2k+1)\pi\) for \(k\in\mathbb{Z}\). So the solution set is exactly \(\{(2k+1)\pi:k\in\mathbb{Z}\}\) (real numbers), matching (B) exactly. Statement (B) is true.

Step 3: Check (C). For \(z=iy\) purely imaginary, \(\cos(iy)=\cosh(y)\), and \(\cosh(y)\to\infty\) as \(y\to\infty\). So \(\cos(z)\) is unbounded on \(\mathbb{C}\), and (C) is false.

Step 4: Check (D). A non-identity Mobius transformation has at most two fixed points. If a Mobius transformation has three or more fixed points, it must fix every point, so it is the identity map \(w=z\), not a constant map (a constant map is not even a valid Mobius transformation, since Mobius transformations must be invertible). So (D) is false as stated.

Step 5: Only (B) is true. \[\boxed{\text{(B)}}\]