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 $u_x = a$?
Best Answer
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
$$f(z)=(ax+b)+i(ay+c)=a(x+iy)+(b+ic)=az+w,$$
which is linear (or affine, depending on how you understand the terms).