Prove that there are no entire function satisfying $|f(z)|\ge |\cos(z)|+|\sin(z)|$ for all $z\in \Bbb C$ Hi. I need to prove that  there are no entire function satisfing  $$|f(z)|\ge|\cos(z)|+|\sin(z)| \\\forall z\in\mathbb{C}.$$
I think I need to use the Liouville theorem.
 Appriciate any help, Thanks!
 A: A hint:
One has $|\cos(x+iy)|^2=\cos^2 x+\sinh^2 y$ and a similar formula for $|\sin(x+iy)|^2$. Use this to obtain an estimate $|f(z)|\geq C>0$.
Now argue about the function $$g(z):={1\over f(z)}\ .$$
A: The identity $\cos^2 z + \sin^2 z = 1$ still holds when $z$ is complex. So always either
$|\cos^2 z| \geq {1 \over 2}$ or $|\sin^2 z| \geq {1 \over 2}$, which in turn implies that
$$|\cos z| + |\sin z| \geq {1 \over \sqrt{2}}$$
So you have an entire function satisfying
$$|f(z)| \geq {1 \over \sqrt{2}}$$
Now use Liouville's theorem in the right way...
A: $$|\cos z|=\left| \frac{e^{iz}+e^{-iz}}{2} \right|=\frac{|z|}2|\cos\alpha|$$
$$|\sin z|=\left| \frac{e^{iz}-e^{-iz}}{2} \right|=\frac{|z|}2|\sin\alpha|$$
where $\alpha=\arg(z)$.
Since $|\cos\alpha|+|\sin\alpha|\geq 1$, we have that $|\cos z|+|\sin z|\geq|z|/2$
Moreover, $|\cos 0|+|\sin 0|=1$, so $|f(z)|\geq1$. There exists some $\delta>0$ such that $|f(z)|>1/2$ for $|z|<\delta$. Let be $C=\min\{\delta/2,1/2\}$.
Therefore, $|f(z)|\geq C$ for all $z\in\Bbb C$, which contradicts strong Picard's theorem. Or, if you prefer, $1/|f|$ is bounded above by $1/C$, and $f$, by Liouville's theorem, is constant, and this is impossible since $|\cos z|+|\sin z|$ is not bounded above (as we have just proven).
