# Complex analytic $f$ with $f(z)=u(x)+iv(y)$ must be linear?

Question:

Let f be a function such that: $$f(x+iy) = u(x) + iv(y)$$ (that is, such that u does not depend on y and v does not depend on $x$). Prove that if f is analytic in C then $f(z) = az + b$ for some a, b elements of C.

Solution:

As f is analytic it satisfies the Cauchy Riemann equations.

$u_x = v_y$, $u_y = -v_x$

By definition $f'(z) = u_x + iv_x$

but $v_x = 0$ so

$f'(z) = u_x$

Now $u_x$ is just some element of C. Call it a.

$f'(z) = a$

Integrate both sides with respect to z and we get

f(z) = az + b

Does this look right? Particularly the part where I let $u_x = a$?

-

Well, initially we must think of $u_x$ as a real variable function, not just some element of $\Bbb C$.
However, from $u_x=v_y$ we can notice the left side is a function of $x$ and the right a function of $y$, while $x$ and $y$ are in-dependent variables, so both sides must in fact be a constant, say $a$. (In fact, this constant must be real, because $u$ and $v$ are real.) Thus $u= ax+b$ and $v=ay+c$ for some constants $b,c$. Writing $w=b+ci$, we have
$$f(z)=(ax+b)+i(ay+c)=a(x+iy)+(b+ic)=az+w,$$