Question

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : The map $f : \mathbb{C} \to \mathbb{C}$ defined by $f(z) = \sin z$ is bounded. Reason R : The function $f(z) = \sin z$ is an entire map.
In the light of the above statements, choose the correct answer from the options given below

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is NOT the correct explanation of A
• C. A is true but R is false
• D. A is false but R is true

Hint:

Do not confuse real trigonometric functions with complex ones! On $\mathbb{R}$, $|\sin x| \leq 1$. On $\mathbb{C}$, $|\sin z|$ grows towards $\infty$ along the imaginary axis!

Answer 1

The Correct Option is D

Step 1: Concept
In complex analysis, trigonometric functions behave differently than in real analysis. While $\sin x$ is bounded on $\mathbb{R}$ ($|\sin x| \leq 1$), $\sin z$ is unbounded on the complex plane $\mathbb{C}$. By Liouville's Theorem, the only bounded entire functions on $\mathbb{C}$ are constant functions.

Step 2: Key Formulas and Approach
The complex sine function is defined as: \[ \sin z = \frac{e^{iz} - e^{-iz}}{2i} \] For purely imaginary numbers $z = iy$ (where $y \in \mathbb{R}$), $\sin(iy) = i \sinh(y)$.

Step 3: Step-by-step Explanation

• Evaluating Reason R: The function $f(z) = \sin z$ is complex-differentiable everywhere on $\mathbb{C}$ with derivative $f'(z) = \cos z$. Thus it is an entire function. Hence, Reason R is true.
• Evaluating Assertion A: Evaluate $|\sin(iy)|$ along the imaginary axis as $y \to \infty$: \[ |\sin(iy)| = |i \sinh(y)| = \sinh(y) = \frac{e^y - e^{-y}}{2} \] As $y \to \infty$, $\lim_{y \to \infty} \sinh(y) = \infty$. Since $|\sin z|$ grows exponentially without bound, $f(z) = \sin z$ is unbounded on $\mathbb{C}$. Hence, Assertion A is false.

Step 4: Final Answer
Assertion A is false, but Reason R is true. Thus, Option (D) is correct.