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$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.

whatanhonor
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3 Answers3

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  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

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

  1. 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 $f(\mathbb{C}) = \mathbb{C}$.

Community
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achille hui
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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!

Conrad
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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 $f(\mathbb C)=\mathbb C$.

Nitin Uniyal
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  • $z(\frac{e^z-1}{z})=e^z-1$ and LHS is a product of an entire onto function and another entire function, while RHS is not onto as it misses $-1$ – Conrad Apr 10 '20 at 01:28
  • $\frac{e^z-1}{z}$ gas removable singularity at $z=0$ hence not entire actually. – Nitin Uniyal Apr 10 '20 at 03:28
  • $\frac{e^z-1}{z}=\sum_{n \ge 0}\frac{z^n}{(n+1)!}$ is an entire function as its Taylor series shows; your claim is incorrect; for any polynomial $P$ one can find an entire function $f$ s.t $Pf$ is not surjective and moreover you can pick any fixed non zero value to miss – Conrad Apr 10 '20 at 04:33
  • even more generally for any entire function $g$ there is an entire function $h$ s.t $gh$ misses a value and again you can pick any fixed non-zero value to miss – Conrad Apr 10 '20 at 04:44
  • @conrad..functions like $\frac{sinz}{z}$ and $\frac{e^z-1}{z}$ are not defined at $z=0$ hence you cannot have a TS unless you define $f(0)=1$ (analytic continuation) in both cases. Both functions have removable singularity at $z=0$. – Nitin Uniyal Apr 10 '20 at 05:11
  • @Conrad.......also see this to distinguish between removable singularity and an entire function....https://math.stackexchange.com/questions/1802420/is-sin-z-z-analytic-at-the-origin – Nitin Uniyal Apr 10 '20 at 05:13
  • $f(z)=\sum_{n \ge 0}\frac{z^n}{(n+1)!}$ is an entire function (check that the Taylor series is defined on the plane); it is easily shown that $zf(z)$ doesn't take value $-1$; quibbling about this or that doesn't make your claim in the post true - it is highly false as noted as it fails for any entire function, not only for polynomials – Conrad Apr 10 '20 at 05:27
  • if for example, you take $f(z)=\sum_{n\ge0}(-\frac{z^{2n}}{(n+1)!})$ which is again a perfectly well defined entire function as its Taylor series converges in the plane, you can check that $z^2f(z)$ misses $1$ – Conrad Apr 10 '20 at 05:33
  • There is difference in the behaviour of $sinz$ and $sinz/z$. The first one is entire while the later has removable singularity at $z=0$. – Nitin Uniyal Apr 10 '20 at 05:34
  • I am just giving you an entire function defined by its Taylor series as above s.t $z^2f(z)$ doesn't take value $1$ - not sure what you mean by your comments; an entire function is defined by its Taylor series - there is no singularity anywhere as the Taylor series converges everywhere - or at least I hope you know this as otherwise, I recommend to go study first some complex analysis as you may be confused about the basics – Conrad Apr 10 '20 at 05:37
  • As per your comments, it seems you are not able to distinguish between removable singularity and entire function. The TS you claimed assumes that $z\ne 0$ as the cancellation of $z$ is performed after expansion of numerator function. Hope it helps you. Now there is no point to carry this discussion further! – Nitin Uniyal Apr 10 '20 at 05:43