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.

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

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

Click 1. $\text{Im} f$ is constant,

2. $f$ is constant,

3. $f \equiv 0$,

4. $f^{'}$ is a non-zero constant.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.****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.**here**for more problems.

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