Іs there any relationship between $v(y)$ and $u(y)$ where $f(iy) = u(y) + iv(y)$?

Question

Let $f(z)$ be a differentiable function of the complex variable $z$ and let $f(iy) = u(y) + iv(y)$ for $z = iy$ where $y$ is a real variable.

Question. Is there any relationship between the real and imaginary parts, $v(y)$ and $u(y)$?

For example, $f(z)$ is unknown, and we are given the explicit expression for $u(y)$, is it possible to find $v(y)$?

I was hoping that there might be some analogue to the Cauchy-Riemann equations, but so far I have not been able to find it.

Answer 1

Quick Answer: No.

For example, let $f_1(z)=z+1$, and $f_2(z)=z^3+1$, both of them are analytic. We get

$$f_1(it)=1+i\cdot t,~~~~f_2(it)=1-i\cdot t^3$$

Both $f_1(it)$ and $f_2(it)$ have the same real part $u(t)=1$, but their imaginary part $v(t)$ are different. Hence, if only given the real function $u(t)$, there is no way to uniquely determine the real function $v(t)$.

---

Remarks:

Let $f(it)=u(t)+i\cdot v(t)$, after taking conjugate on both sides, we get $f(-it)=u(t)-i\cdot v(t)$. Take the addition and subtraction, we get

$$u(t)=\frac{f(it)+f(-it)}2,~~~~v(t)=\frac{f(it)-f(-it)}{2i}$$

We can see

$u(t)=u(-t)$, implies $u(t)$ is even function,

$v(-t)=-v(t)$, implies $v(t)$ is odd function.

Answer 2

Even if $f$ is holomorphic, $u$ and $v$ can be any real analytic functions (and they are always real analytic because $f$ is), independant one from the other. Indeed, if $u$ and $v$ are real analytic on some open interval $I \subset \mathbb{R}$, take some $a \in I$ and write $u(y) = \sum_{n \geqslant 0} \alpha_n(y - a)^n$ and $v(y) = \sum_{n \geqslant 0} \beta_n(y - a)^n$ in some interval $]a - r,a + r[$. $u$ and $v$ are real thus each $\alpha_n$ and $\beta_n$ are real.

Therefore, you can set $f_a(z) = \sum_{n \geqslant 0} (\alpha_n + i\beta_n)(-iz - a)^n$, which is well defined on $U_a = -iB(a,r)$. We have $f_a(iy) = u(y) + iv(y)$ when $y \in \mathbb{R} \cap iU_a = ]a - r,a + r[$. If $U_a \cap U_b \neq \emptyset$, we have $f_a(iy) = f_b(iy) = u(y) + iv(y)$ when $y \in -iU_a \cap -iU_b \cap \mathbb{R}$, which is open and non-empty in $\mathbb{R}$ (so it admits accumulation points). By uniqueness of the analytic continuation, $f_a = f_b$ on $U_a \cap U_b$.

Therefore, the $(f_a)$ stick together to make a unique holomorphic function $f$ well defined on $U = \bigcup_{a \in I} U_a$, which is an open neighborhood of $-iI$ in $\mathbb{C}$ and $f$ verifies $f(iy) = u(y) + iv(y)$ when $y \in I$. It proves that such a $f$ always exist under the only condition that $u$ and $v$ both are real analytic.