Range of entire function. Picard's theorem

Question

$f(z)$$=$ 2$z^{14}$$cos^2z$

$f(z)$ is entire function. I want to figure out if $f$($\Bbb C$)$=$$\Bbb C$ or there's a point $a$ such that $f$($\Bbb C$)$=$$\Bbb C$\ {$a$}. So essentially I want to use Little Picard Theorem here.

I thought about using the reflection principle. Let $f(z)$$=$$w$. So here $f$($\bar{z}$)=$\bar{f(z)}$. This means that if $w$ if this exceptional point then $\bar{w}$ is too. I guess this implies that exceptional points can only be located on the real line?

Can someone please correct me if I am wrong. Any help with solving this problem would be appreciated.

Answer 1

1. Along the real axis, we have $f(z) = f(x) = 2x^{14}\cos(x)^2 \ge 0$.

   Notice $f(0) = 0$ and $\lim\limits_{n\to\infty} f(n\pi) = 2(n\pi)^{14} \to \infty$. Apply IVT along positive real axis, we have

$$f([0,\infty)) \supset [0,\infty)\quad\implies\quad f(\mathbb{R}) = [0,\infty)$$

2. Along the imaginary axis, we have $f(z) = f(iy) = -2y^{14}\cosh(y)^2 \le 0$.

   Notice $f(i0) = 0$ and $\lim\limits_{y\to+\infty} f(iy) = -\infty$. Apply IVT along positive imaginary axis, we have $$f(i[0,\infty)) \supset (-\infty,0]\quad\implies\quad f(i\mathbb{R}) = (-\infty,0]$$

Combine them, we have $f(\mathbb{R} \cup i\mathbb{R}) = \mathbb{R}$.

As you have already known, by reflection principle, if $f(z)$ avoids any number $\alpha$, $\alpha$ need to be real. Since this have been ruled out by above argument, $f(z)$ does not avoid any number and $f(\mathbb{C}) = \mathbb{C}$.

Answer 2

Assume there is $a \in \mathbb C, f(z)-a$ has no roots.

$h(z)=\log(f(z)-a)$ is an entire function and since $|f(z)| \le AR^{14}e^{2R}, |z|=R$ large enough where $A>0$ is a fixed constant, it follows that $\log |f(z)-a|=\Re h(z) \le 3R, |z|=R$ for all $R \ge R_0$ large enough, so applying Borel Caratheodory with $R,2R, R \ge R_0$, we get $|h(z)| \le 6R+3|h(0)|, |z|=R\ge R_0$ which immediately implies that $h(z)=bz+c$ for some complex $b,c$ and then we get:

$2z^{14}\cos^2z=a+e^ce^{bz}$

Putting $z=0$ we get $a+e^c=0$, so $2z^{14}\cos^2z=-a(e^{bz}-1)$ and now obviously LHS has a zero of order $14$ at zero, while RHS is either identically zero or has a simple zero at zero, so either way, we get a contradiction and we are done!

Answer 3

Notice that $f(z)$ is the product of two entire functions $g(z)=2z^{14}$ and $h(z)=cos^2z$ where $g$ is onto by fundamental theorem of algebra, so $f(\mathbb C)=\mathbb C$.