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would any one tell me: which of the function is entire $\sin(\sqrt{z})$ and $\cos(\sqrt{z})$. I have no Idea in which logic I need to apply here. Thank you.

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Have you tried looking at the Maclaurin series of these two functions? –  J. M. Aug 4 '12 at 17:06
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Morally: the "function" $z \mapsto \sqrt{z}$ fails to be entire because of the ambiguity in the square root: for any $z$ there are two choices of square root, and there's no continuous way to make a consistent choice on the whole complex plane. When you compute $\sin(\sqrt{z})$, does it depend on which choice of square root you made? What about $\cos(\sqrt{z})$? –  user29743 Aug 4 '12 at 17:12
    
Thank you....... –  miosaki Aug 4 '12 at 19:51
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1 Answer

up vote 1 down vote accepted

Although countinghaus gave a nice informal answer in comments, I'll add some formal details: $$\cos \sqrt{z} = \sum_{n=0}^\infty (-1)^n\frac{z^n}{(2n)!} \tag1$$ is an entire function. So is $$\frac{\sin \sqrt{z}}{\sqrt{z}} = \sum_{n=0}^\infty (-1)^n\frac{z^n}{(2n+1)!} \tag2$$ but not $\sin\sqrt{z}$ itself. One can argue from $\sqrt{z}$ being two-valued, but there's another way: since $\sin\sqrt{x}$ does not have a one-sided derivative at $x=0+$, it cannot be extended to an entire function.

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