Question:
a) Let $f(z) = u(x,y) + iv(x,y)$ be analytic in $\Omega$, suppose that $v(x,y) = e^{-y}(y\cos x -x \sin x)$. Find $f(z)$.
b) Let $f(z), g(z)$ be analytic in an open, connected domain $\Omega$ containing $0$, and suppose that
$$f(z+w) = f(z)f(w) -g(z)g(w), \ \ \ g(z+w) = g(z)f(w)+g(w)f(z),$$
whenever $z,w$, and $z+w$ are all contained in $\Omega$. Suppose further that $f(0) = 1$, $f'(0) = 0$, find all such pairs $f(z)$, $g(z)$.
My attempt:
a) This seems like it should be an application of the Cauchy-Riemann equations, but I'm struggling with it. So, we have
$$\frac{\partial v}{\partial y} = -e^{-y}(y\cos x-x\sin x) + e^{-y}\cos x$$ $$=e^{-y}\left[(1-y)\cos x + x\sin x\right],\ \ \ \text{ and }$$
$$\frac{\partial v}{\partial x} = e^{-y}(-y\sin x -\sin x -x\cos x)$$ $$=-e^{-y}\left[(1+y)\sin x + x\cos x\right].$$
Thus, we have
$$\frac{\partial u}{\partial x} = e^{-y}\left[(1-y)\cos x + x\sin x\right], \ \ \ \text{ and }$$ $$\frac{\partial u}{\partial y} = e^{-y}\left[(1+y)\sin x + x\cos x\right].$$
Now, we can integrate, to get
$$u(x,y) = -e^{-y}\left[(y-2)\sin x + x\cos x\right] + G(y),$$ $$u(x,y) = -e^{-y}\left[(y+2)\sin x+x\cos x\right] + H(x).$$
But, setting these equations equal to each other, I'm getting
$$4e^{-y}\sin x= H(x)-G(y),$$
and I don't know how to solve for $H(x)$ and $G(y)$ from here.
b) Since part a) uses the Cauchy-Riemann, I was thinking this part would, but I haven't quite figured out have to apply them yet. All I've really managed to find so far is that
$$f(0) = 1 \implies g(0) = 0,$$
unless $g(z) = 0$ for all $z \in \Omega$. So, I guess one pair is $f(z)$ analytic, and $g(z) = 0$. But, that can't be all of the pairs!
Hints and ideas of where to go from here would be much appreciated, on either or both of a) and b)
Regarding part b), can you spot a pattern? I.e. don't you know a pair of functions over $\mathbb{R}$ or $\mathbb{C}$ which satisfies that kind of thing?
You need to apply the pattern detected in two different ways, of course, for you have two unknown functions. Any idea to start with?
Anyway, you can play with the two variables $z,w$ and, if you differentiate with respect to one and assume the other is constant, you can figure out e.g. the value $g(0)$. Have you tried that? You might even obtain some kind of differential equation which will give you the shape of $f,g$.
As for part a), I like to do the following with this kind of problems. You're on the whole complex domain, which is simply connected (should the domain be punctured, you need to be extra careful when integrating). Both components $u, v$ of $f$ must be harmonic, i.e.
$$\Delta u=\Delta v=0.$$
This is a necessary condition, in general. Our function $v$ thus must be harmonic to begin with (if not, this means that the exercise is ill-posed).
If we use the operators $\partial, \overline{\partial}$ this is written as follows.
The Cauchy-Riemann equations for $f$ are equivalent to saying that
$$\overline{\partial}f=0.$$
Instead of computing $f$ directly, we may try and obtain, say, $\partial v$, but you need to relate it to the original $f$, it should give you a constant multiple of $f'.$ Look for information on the operators $\partial, \overline{\partial}.$
The fact is that, if $v$ is harmonic, then $\overline{\partial}\left(\partial v\right)=0$, i.e. $\partial v$ is holomorphic.
A good reference is G. Polya, G. Latta, Complex Variables, but any textbook will do.