the range of the entire function $ e^z(1+\cos\sqrt{z} ) $ Picard's Great theorem for

Question

Let $$f(z) = e^z (1+\cos\sqrt{z} ) $$ $\Omega=\{z\in\Bbb C: |z|\gt r\}$, $r\gt 0$. What is $f(\Omega)$?

where $\sqrt{z}=\exp{(\text{Log }z/2)}, \text{Log }z=\log|z|+i\arg z,\arg z\in(-\pi,\pi]$

$\infty$ is an essential singularity. Picard's Great theorem , $\Bbb C\setminus f(\Omega) $ contains at most one point. $f(\Omega)$ is $\Bbb C$ ?

When $ z\in\Bbb R $, $f(z)\geq0$. According to Schwarz reflection principle and Picard's Great theorem, all $ x+iy(y\ne0), \in\Bbb f(\Omega) $. How to show that all negative real numbers belong to $f(\Omega)$? thanks a lot!

Is there a simple, more Elementary Proof than Conard's ?

Answer 1

Claim $f_a(z):=e^{z}(1+\cos\sqrt z)-a$ has infinitely many roots for any $a\in\mathbb C\setminus\{0\}$.

(Clearly the case $a=0$ is trivial. For simplicity I excluded this special case.)

Proof: To prove this claim by contradiction, the following lemma is used:

Lemma $F_a:=\frac{f'_a}{f_a}$, $R=[-\frac\pi 2,\frac\pi2]\setminus\{0\}$ $$\lim_{N\to\infty}F_a(R_N e^{i\theta})= \begin{cases} 1, && \theta\in R \\ 0, && \theta\in[-\pi,\pi)\setminus R \end{cases} $$ where $R_N=(2N+1)^2\pi^2$.

Suppose otherwise, i.e. $f_a$ has finitely many roots, then by argument principle $$\oint_{|z|=R_N}F_a(z)dz=2\pi iK$$ where $K$ is constant for sufficiently large $N$. (In addition, under this hypothesis, no singularities lie on the integration path for sufficiently large $N$, and thus the integral is well-defined.)

By parametrization, $$\begin{align} \int_{-\pi}^{\pi}F_a(R_Ne^{i\theta})R_Nie^{i\theta}d\theta&=2\pi i K \\ \int_{-\pi}^{\pi}F_a(R_Ne^{i\theta})e^{i\theta}d\theta &=\frac{2\pi K}{R_N} \\ \lim_{N\to\infty}\int_{-\pi}^{\pi}F_a(R_Ne^{i\theta})e^{i\theta}d\theta &= \lim_{N\to\infty}\frac{2\pi K}{R_N}\\ \int_{-\pi}^{\pi}\lim_{N\to\infty}F_a(R_Ne^{i\theta})e^{i\theta}d\theta &=0 \qquad \qquad \qquad (\star)\\ \int_{-\pi/2}^{\pi/2}e^{i\theta}d\theta &= 0 \\ 2&=0 \end{align} $$ (($\star$): the interchange of integration and limit is justified by Dominated Convergence Theorem.)

Contradiction.

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Sketch of proof of lemma: $$F_a(z)=\frac{1+\sec\sqrt z-\frac{\tan\sqrt z}{2\sqrt z}}{1+\sec\sqrt z-ae^{-z}\sec\sqrt z}$$

Exploiting the estimates $$\sec\sqrt{Re^{i\theta}}=\frac{2}{\exp(i\sqrt R e^{i\theta/2})+\exp(-i\sqrt R e^{i\theta/2})}=O\left(e^{-|\sin\frac{\theta}{2}|\sqrt R}\right)\quad \theta\ne0$$ $$\frac{\tan\sqrt {Re^{i\theta}}}{\sqrt {Re^{i\theta}}}=O\left(\frac1{\sqrt R}\right)\qquad\theta\ne 0$$ $$e^{-Re^{i\theta}}=O\left(e^{-R\cos\theta}\right)$$

immediately proves the lemma for $\theta\ne 0$.

Note that when $\theta=0$, $F_a(R_N)=\frac{1-1-0}{1-1-ae^{-R_N}}=0$.$\blacksquare$

Answer 2

Below we prove that $f(z)=a$ has infinitely many solutions for any complex $a$, hence there are solutions with arbitrarily large modulus, so the image required is the full plane.

Assume by contradiction there is an $a$ s.t. $f(z)=a$ has only finitely many solutions. Then since $f$ is entire of order $1$ and type $1$ since $\cos{\sqrt z}$ has order $\frac{1}{2}$ and $e^z$ has obviously order $1$, type $1$, it immediately follows that $f(z)-a=e^zP(z)$ for some polynomial $P$

(proof of the claim above: eliminating the finitely many zeroes by dividing by an appropriate $P$ with same zeroes, we get that $f(z)-a=e^{Q(z)}P(z)$ and then by order theory $Q$ must be a linear polynomial $\alpha z+b$, then by type $|\alpha|=1$ and then $\alpha=1$ by looking at the positive axis behavior and we can absorb $e^b$ into $P$)

But now we get $1+\cos{\sqrt z}-P(z)=ae^{-z}$ and that is obviously impossible since the LHS is an entire function of order $\frac{1}{2}$ while the right hand side has order $1$ since plainly $a\ne 0$

(if one is not comfortable with order/type for entire functions, just look at $\frac{\log M(R)}{R}$ for large $R$, where $M(R)$ is the supremum on the $R$ disk of the various functions involved, so for example for $f-a$, clearly $\log M_{f-a}(R)$~$R$ asymptotically for most - we can have $\cos \sqrt R =-1$ of course - large $R$, so $M_Q(R)$ must be same which immediately implies $Q$ linear polynomial with leading term $|\alpha|=1$, but then noting that on the positive real axis we get $\log \frac{(f-a)(R)}{P(R)}$~$R$, for most $R$'s $\alpha=1$ etc)