Is $f(x + iy) = \exp(iy)$ analytic?

Question

I know that if $|f| = c$ and $f$ is analytic in $D$ then $f = k$ in $D$.

If I take $f(z) = f(x + iy) = \exp(iy)$ then $|f| = 1$ in the complex plain.

But $f(z)$ is not constant, does this means that $f$ is not analytic? And if so why not?

Answer 1

Indeed it is not analytic. If you write $f(x+yi)$ as $u(x,y)+v(x,y)i$, with $u(x,y),v(x,y)\in\Bbb R$, then $u(x,y)=\cos(y)$ and $v(x,y)=\sin(y)$. But then both $u_x$ and $v_x$ are the null function. On the other hand $u_y(x,y)=-\sin(y)$, and $v_y(x,y)=\cos(y)$. So, no point of $\Bbb R^2$ is a solution of the Cauchy-Riemann equations, and therefore $f$ is nowhere complex-differentable. Therefore, it is not holomorphic, and, in particular, it is not analytic.

Answer 2

Another reason: $|f(z)| = 1$ so $f$ is bounded. Thus $f$ is not analytic by Liouville's theorem