Complex-valued functions with value and derivatives having absolute less than 1 everywhere: Does $f(0)=|f'(0)|=1$ imply $f(t)=e^{\pm it}$? 
Are the functions $f(t)=e^{\pm it}=\cos(t)+i\sin(t)$ the only functions $f:\Bbb R \to \Bbb C$ such that $|f^{(n)}(t)|\leq 1$ for all $n=0,1,\ldots$ and $t\in \Bbb R$ such that $f(0)=|f'(0)|=1$?

It has been shown that $g(t)=\sin(t)$ is the unique real-valued function with value and all its derivatives less than 1 in absolute value and $g'(0)=1$: If $f$ is a smooth real valued function on real line such that $f'(0)=1$ and $|f^{(n)} (x)|$ is uniformly bounded by $1$ , then $f(x)=\sin x$?
Since the function $g(t)=\mathrm {re} (f(t)/f'(0))$ satisfies the specified conditions, it follows that the real part of the function $f/f'(0)$ must be equal to $\sin(t)$. In particular, the real part of $f(0)/f'(0) = 1/f'(0)$ is 0, which means that $f'(0)$ is imaginary, and so $f'(0)=\pm i$, so that $|f'(0)|=1$. We conclude that the imaginary part of $f$ is $\pm i \sin(t)$.
Regarding the real part of $f$ we know that it equals 1 at $t=0$ and it needs to be 0 at $t=\pi/2$ to ensure that $|f(\pi/2)| \leq 1$. Does this imply that the real part of $f$ is $\cos(t)$?
 A: This actually follows directly from Bernstein's theorem (the extension of the well-known trigonometric polynomials inequality between the norm of the derivative and the norm of the function) which says (in this context) that if $f$ entire, bounded on the real axis $|f(x)| \le M$ and $|f(z)| \le Ce^{|z|}$ for some $C>0$ then $|f'(x)| \le M, x \in \mathbb R$ and if we have equality at a point, then $f(z)=ae^{iz}+be^{-iz}, |a|+|b|=1$
For a simple self-contained proof using just basic complex analysis (Phragmen Lindelof) see page 513 of the book Analytic Theory of Polynomials by Q Rahman and G Schmeisser
The hypothesis on $f$ (and the only place we use the full derivatives bounds for all $x$) implies that $f$ is entire. Then if $f(z)=\sum a_nz^n, a_n=\frac{f^{(n)}(0)}{n!}$ so the bounds at $0$ only, imply $|f(z)| \le \sum |z^n|/n! =e^{|z|}$ so by Bernstein's theorem we get that since $|f'(0)|=1$ then $f(z)=ae^{iz}+be^{-iz}, |a|+|b|=1$; using $f(0)=1$ we get $a+b=1=|a|+|b|$ so $a,b \ge 0$ and then $|f'(0)|=1$ implies $|a-b|=1$ hence $a=1, b=0$ or $a=0, b=1$ and we are done!
Note that the proof shows that what we need is only that $|f(x)| \le 1, f(0)=1, |f'(0)|=1, |f^{(n)}(0)|\le 1$ and $f$ is analytic on the real line (meaning the Taylor series at any point converges to $f$ in a small interval centered at the point), which for example is implied by much weaker bounds on the derivatives.
Then writing the Taylor series of $f$ at $0$ the condition $|f^{(n)}(0)|\le 1$ implies that it defines an entire function equal to $f$ in a neighborhood of $0$ and since $f$ is real analytic on the real line, it equals $f$ there by the identity principle and then everything goes as before
