# Evaluate a complex set

$$E_{R}=\{\cos{z} | z \in \mathbb{C}, |z|>R\}$$

For any $0<R \in \mathbb{R}$.

My feeling is the $E_R = \mathbb{C}$, for any $R$, but I don't know how to show it, if it's true.

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Do you know the answer for $e^z$? Do you know Picard theorem? –  Fabian Jan 22 '12 at 9:39
Now that you've mentioned it, yes, Picard theorem can show that it's the complex plane minus the origin. You should post it as an answer, so I can mark it as the accepted answer. –  ofer Jan 22 '12 at 9:53
Note that $\cos$ assumes the value $w=0$! –  Fabian Jan 22 '12 at 11:37

The function $\cos$ is periodic with period $2\pi$; therefore any vertical strip of width $2\pi$ in the $z$-plane, $z=x+iy$, will give rise to the full set of values of $\cos$. By excluding the $z$'s with $|z|\leq R$ there is an infinity of such strips left, so there are no values excluded. It follows that we may as well look at the values of $\cos$ in the strip $0\leq x\leq 2\pi$. Now $$\cos z=\cosh y \ \cos x - i\ \sinh y\ \sin x\ .$$ Keeping $y\geq0$ fixed and letting $x$ go from $0$ to $2\pi$ the point $w:=\cos z$ describes an ellipse with horizontal semiaxis $\cosh y$ and vertical semiaxis $\sinh y$, and it is easily seen that the family of these ellipses covers all of ${\mathbb C}$.

In this very special example we needed neither the theorem of Casorati-Weierstrass nor the really deep Picard's theorem, but of course the function $\cos$ may serve as an explicit instance for these two theorems.

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A different approach. Let $w\in\mathbb{C}$. We must find $z\in\mathbb{C}$ such that $|z|>R$ and $$\cos z=\frac{e^{iz}+e^{-iz}}{2}=w\implies e^{2iz}-2\,w\,e^{iz}+1=0\ .$$ Solving for $e^{iz}$ we obtain $$e^{iz}=w\pm\sqrt{w^2-1}\ .$$ One at least of $w+\sqrt{w^2-1}$ or $w-\sqrt{w^2-1}$ is non-zero. Assume $w+\sqrt{w^2-1}\ne0$. Then $$z=\frac1i\,\log(w+\sqrt{w^2-1})+2\,k\,\pi ,\quad k\in\mathbb{Z},$$ where $\log$ is the principal branch of the logarithm. We can take $k$ large enough to have $|z|>R$.

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Hint: Make use of Picard theorem to show that it the image is all of $\mathbb{C}$ with at most a single point missing.

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But, of course, Picard is far more difficult than the result requested here... –  GEdgar Jan 22 '12 at 19:32
From Picard theorem I know that this set is or the whole complex plane, or the whole complex plane minus one point. How can decide which of the following is our case? –  ofer Jan 23 '12 at 20:11