How to show the existence of an entire function

Question

I have been working on this problem for quite sometime.

For part (i), I obtained the Taylor series for $4\sin(z) - \sin(4z)$.

At $z = -\pi$, the Taylor series is:

$4\sum_{n=0}^{n} \frac{(z + \pi)^{2n+1}}{(2n+1)!}$ - $\sum_{n=0}^{n} \frac{4^{2n+1}(4z+\pi)^{2n+1}}{(2n+1)!}$

I am actually quite confused about part (ii).

It seems quite obvious to me but I am not sure how to show the existence of the entire function g(z).

$4sin(z) + sin(4z)$ is definitely entire for all $z \in C $ but $z \neq -\pi$

Since $z + \pi \neq 0$ $\in C $ $\setminus \{- \pi \}$, g(z) is entire.

Can this be a valid proof? How can I do better ? I'll appreciate if someone can advise me.

Thanks

Answer 1

Note that $\sin(z)=-\sin(z+π)$ and $\sin(4z)=\sin(4z+4π)=\sin(4(z+π))$, so that the signs for the Taylor series are actually the opposite.

Now you need to compare the terms for $n=0$ and recognize that they cancel, so that the leading terms are of third order.

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Alternatively $$ \sin(4z)=4\sin(z)\cos(z)\cos(2z) $$ so with $w=z+π$ $$ 4\sin(z)+\sin(4z)=-4\sin(w)(1-\cos(w)\cos(2w))\\ =-4\sin(w)·(1-(1-2\sin^2(w/2))(1-2\sin^2(w))) $$ etc. leads to an expression of $\sin^3(w/2)$ times an entire function and $\sin(w/2)/w$ is an entire function.