### CSIR JUNE 2011 PART C QUESTION 80 SOLUTION ($\text{Re}(f(z)$ bounded then $f(z)$ is constant: Three different proofs)

Let $f$ be an entire function. If $\text{Re} f$ is bounded then
1. $\text{Im} f$ is constant,
2. $f$ is constant,
3. $f \equiv 0$,
4. $f^{'}$ is a non-zero constant.

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Solution:
First, we prove that option 2 is correct which will imply that option 1 is also true.

Option 2: (True)
Proof 1: Given that $f$ is entire. Assume $f$ is not a constant function, then by little Picards's theorem, range of $f$ is either $\Bbb C$ or $\Bbb C \backslash \{a\}$ for some $a \in \Bbb C$. Therefore $\text{Re}(f)$ is unbounded. It is given that $\text{Re}(f)$ is bounded. Hence $f$ has to a constant function.
Proof 2:
Assume that the real part of $f$ is bounded then there exists $k$ such that $\text{Re}(f(z)) \le k$. We define $h(z) = \frac{1}{k - f(z) + 1}$. Now $\text{Re}f(z) \le k$ implies that $\text{Re} (k - f(z)) \ge 0$ and $\text{Re}(k-f(z)+1) \ge 1$. This shows that the denominator $k-f(z)+1$ will never zero and hence $h(z)$ is an entire function. Let $h(z) = \frac{1}{u + i v} = \frac{u-iv}{u^2 + v^2} = \frac{u}{u^2+v^2} -i \frac{v}{u^2 + v^2}$. We have $u = \text{Re}(k-f(z)+1)$ and $|h(z)|^2 = (\frac{u}{u^2+v^2})^2+(\frac{v}{u^2+v^2})^2 = \frac{1}{u^2 + v^2} \le 1$ since $u = \text{Re}(k-f(z)+1) \ge 1$. This shows that $|h(z)| \le 1$ and hence h(z) is a bounded entire function which has to be constant by the Liouville's theorem. But $h(z)$ is constant implies that $f(z)$ is constant.
proof 3: Let $f(z) = u + iv$. It is given that $u$ is bounded. Consider the function $h(z) = e^{f(z)}$. Now, $|e^z| = |e^{u+iv}| = |e^u| \,|e^{iv}| = e^u$ which is bounded as $u$ is bounded. By Liouville's theorem $h(z)$ is constant and hence $f(z)$ is also constant.
option 4:(False)  We have proves that $f$ is constant and hence its derivative $f^{'}$ has to be zero.
option 1: (True) We have shown that the function $f$ itsels is constant in the previous option. Hence $\text{Im}(f)$ has to be bounded.
option 3:(False) Consider the constant function $f(z) = 1$ for all $z \in \Bbb C$. Then $\text{Re}(f)$ is bounded but $f$ is not identically zero.

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