= Solution
<Rouché's theorem> states that if $f,g$ are holomorphic on and inside a simple closed contour and $|g|<|f|$ on it, then $f$ and $f+g$ have the same number of zeros inside, counted with multiplicity. On $|z|=1$, the <power series> for sine gives
$$
|\sin z|\leq\sum_{j=0}^\infty\frac1{(2j+1)!}=\sinh1<2=|2z|.
$$
Apply <Rouché's theorem> with $f=2z$ and $g=-\sin z$. There is one zero inside and none on the boundary. Since $z=0$ is a zero and $(2z-\sin z)'|_{z=0}=1$, it is simple. \b[Exactly one complex number qualifies: $z=0$.]
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