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

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?

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).
