Question

Which of the following function is analytic on $\mathbb{C}$

• A. $f(z) = \sin(z^2)$
• B. $f(z) = \frac{1}{z}$
• C. $f(z) = |z|^2$
• D. $f(z) = e^{\bar{z}}$

Hint:

Standard entire functions include: polynomials $P(z)$, $e^z$, $\sin z$, $\cos z$, $\sinh z$, $\cosh z$. Any finite combination or composition of these is also entire!

Answer 1

The Correct Option is A

Step 1: Concept
A function $f(z)$ is said to be analytic on $\mathbb{C$} (or an entire function) if it is complex-differentiable at every point in the complex plane. The composition of two entire functions is always an entire function.

Step 2: Key Formulas and Approach
We analyze each function for singularities or non-differentiability across $\mathbb{C}$.

Step 3: Step-by-step Explanation

• Testing Option (A): $f(z) = \sin(z^2)$
The polynomial $g(z) = z^2$ is entire. The sine function $h(w) = \sin w$ is entire. The composition $f(z) = h(g(z)) = \sin(z^2)$ is therefore differentiable everywhere on $\mathbb{C}$. Its derivative is $f'(z) = 2z \cos(z^2)$, which exists everywhere. Hence, $f(z)$ is analytic on $\mathbb{C}$.

• Testing Option (B): $f(z) = \frac{1{z}$}
This function has a pole (isolated singularity) at $z = 0$. Hence, it is not analytic on all of $\mathbb{C}$ (it is only analytic on $\mathbb{C} \setminus \{0\}$).

• Testing Option (C): $f(z) = |z|^2 = x^2 + y^2$
Here $u = x^2 + y^2$ and $v = 0$. C-R equations give $2x = 0$ and $2y = 0$, which holds only at $z = 0$. Since it is differentiable at a single isolated point, it is not analytic anywhere.

• Testing Option (D): $f(z) = e^{\bar{z = e^{x - iy}$}
Here $\frac{\partial f}{\partial \bar{z}} = e^{\bar{z}} \neq 0$. Functions depending on $\bar{z}$ fail C-R equations everywhere and are nowhere analytic.

Step 4: Final Answer
The function $f(z) = \sin(z^2)$ is entire (analytic on $\mathbb{C}$). Thus, Option (A) is correct.