Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 4 / a

The identity theorem says that two holomorphic functions on a connected domain which agree on a set with an accumulation point inside that domain agree everywhere there.

No such entire function exists. On the subsequence $1/(2m)\to 0$ indexed by even numbers, the proposed function equals $\sin z$. The identity theorem would force $f(z)=\sin z$ throughout $\mathbb{C}$. At the points indexed by odd numbers it is instead required to equal $-\sin z$, a contradiction since $\sin (1/n)>0$ for positive $n$. The accumulation point zero lies inside the assumed domain of holomorphy, which is essential to this argument.

Yes: take the unnormalized sinc function with its removable value at zero,Its power series is ${\sum }_{m=0}^{\infty }(-1{)}^m{z}^{2m}/(2m+1)!$, with infinite radius of convergence, so it is an entire function. At $z=1/n$ it gives $n\sin (1/n)$ as required. The identity theorem also proves this entire solution is unique.

No such entire function exists. If it did, the entire function $h(z{)}^2-z$ would vanish at the distinct points $(n/(n+1){)}^2$, whose accumulation point is one. The identity theorem would give $h(z{)}^2=z$ on all of $\mathbb{C}$.

At zero this forces $h(0)=0$. Differentiating the identity gives $2h(z)h^{\prime }(z)=1$, impossible at zero. Equivalently, a squared holomorphic function has an even-order zero, whereas $z$ has a simple zero. This is the obstruction expressed by no entire square root of the identity function; a square-root branch near one cannot extend holomorphically through zero.