Step 1: Understanding the Problem:
This question asks us to identify which of the given complex-valued functions is analytic (entire) over the entire complex plane $\mathbb{C}$.
Step 2: Key Formula or Approach:
A function $f(z)$ is analytic over the entire complex plane (entire) if it is differentiable at every point in $\mathbb{C}$.
A function is not entire if it contains points of singularity (such as poles, essential singularities, or branch cuts).
Step 3: Detailed Explanation:
• Let us analyze each of the given functions to determine its analyticity:
• Option A: $f(z) = e^{\frac{1}{z}}$.
The function $\frac{1}{z}$ is not defined at $z = 0$.
As $z \to 0$, the function exhibits an essential singularity. Because it is not differentiable at $z = 0$, it is not analytic over the entire complex plane.
• Option B: $f(z) = \cos z$.
The complex cosine function is defined in terms of exponential functions:
\[ \cos z = \frac{e^{iz} + e^{-iz}}{2} \]
Since $e^z$ is an entire function, any composition of $e^z$ with linear functions (like $iz$ and $-iz$) is also entire.
The sum of entire functions is also entire. Therefore, $\cos z$ is analytic at every point in the complex plane.
• Option C: $f(z) = \frac{2}{1-z}$.
This rational function has a singularity where the denominator is zero:
\[ 1 - z = 0 \implies z = 1 \]
At $z = 1$, the function has a simple pole, making it non-analytic at this point.
• Option D: $f(z) = \ln z$.
The complex logarithm function has a branch point at the origin $z = 0$ and requires a branch cut (typically along the negative real axis) to be defined as a single-valued function.
It is not continuous or differentiable across the branch cut, so it is not analytic over the entire complex plane.
Step 4: Final Answer
The function $\cos z$ is the only function analytic over the entire complex plane, which corresponds to option (B).