Can we use sine and cosine to express this complex (entire) function in terms of $z$?

Question

I believe $f: \mathbb C \to \mathbb C, f(z)=u+iv=e^x(a(x,y)+ib(x,y))$, with $u,v: \mathbb R^2 \to \mathbb R$,

$$a(x,y)=(x+1)\cos(y)-y\sin(y), u(x,y)=e^xa(x,y)$$

$$b(x,y)=(x+1)\sin(y)+y\cos(y), v(x,y)=e^xb(x,y)$$

is entire.

I'm supposed to write this in terms of $z$. Besides doing the obvious $(x,y)=(\Re(z),\Im(z))$, my idea is to do something like

$$f(z) = e^x |z+1|e^{i\theta(z)}$$

where

$$\cos(\theta(z)) = a(x,y) \frac{1}{|z+1|} = [(x+1)\cos(y)-y\sin(y)] \frac{1}{|z+1|}$$

$$\sin(\theta(z)) = b(x,y) \frac{1}{|z+1|} = [(x+1)\sin(y)+y\cos(y)] \frac{1}{|z+1|}$$

But what's my $\theta$?

If we're talking about just $\mathbb R$, then I think we can just have as follows:

For $f,g: \mathbb R \to \mathbb R$ s.t. $f^2+g^2=1$, we can express (reparametrise?) $f = \cos \circ \theta$ and $g = \sin \circ \theta$. Here, $\arccos \circ f = \arcsin \circ g$ and then $\theta: \mathbb R \to \mathbb R$ is uniquely $\theta=\arccos \circ f = \arcsin \circ g$.

I seem to recall there isn't like a complex $\arcsin$ or $\arccos$

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Edit: Ok so it's been answered with $f(z) = (1+z) e^z$, but there isn't really an explanation as to how to do this in general.

I mean to ask like how we think of $x^2+2x = (x+1)^2-1$. There, the rule is completing the square $x^2+px = x^2+px + (\frac{p}{2})^2 - (\frac{p}{2})^2$.

Here, what's the rule?

Answer 1

Just take the function $f(z) = (1+z) e^z$

I found this solution recognizing the real and imaginary parts. But then I learned about the Milne-Thomson method which allows you to find $f$ if you know the real and the imaginary parts (see the example 1 on the link).

Answer 2

Based on jjagmath's link Milne-Thomson method for finding a holomorphic function, it seems we have like

'$f(z)=u(z,0)+iv(z,0)$'

as in for real $u,v$ consider their (unique?) complex extensions $\tilde u, \tilde v: \mathbb C^2 \to \mathbb C$, where $\tilde u(z,w)=u(z,w)$ for all $(z,w) \in \mathbb R^2$ (and similar for $v$) (treating $\mathbb R$ as a literal subset of $\mathbb C$ and $\mathbb R^2$ as a literal subset of $\mathbb C^2$).

The proof appears very elementary using just $(x,y)=\frac{1}{2}(z+\overline z, -i(z-\overline z))$, but I think there's more to think about re 'This is an identity even when x and y are not real' when you think of $\tilde u$ instead of $u$ (and similar for $v$)

But anyway, as a heuristic Ansatz or whatever pretending $u$ is the same as $\tilde u$ in our case we have

$$u(z,0)=e^z(z+1), v(z,0)=0$$

Hence,

$$f(z)=e^z(z+1)+0i=e^z(z+1)$$