# Proving an equation relating to the Cauchy-Riemann Equations.

Here is what the question asks:

Let $$\Omega \subset \mathbb{C}$$ be a domain (meaning that it's a connected and open), and let $$f: \Omega \rightarrow \mathbb{C}$$ be holomorphic. Prove that if $$f(z) = iv(x, y)$$ for every $$z \in \Omega$$ (namely, if the real part of the function is $$u \equiv 0$$) then there is a constant $$v_0 \in \mathbb{R}$$ such that $$f(z) ≡ iv_0 \in \Omega$$

I have a vivid understanding of this problem, but I am not sure what the "$$\equiv$$" implies $$u \equiv 0$$ and $$f(z) ≡ iv_0$$.

Although, here is how I went about solving it:

For the function to be Complex Differentiable, we know that the Cauchy-Reimann equations must be satisfied.

Let $$u(x,y)$$ and $$v(x,y)$$ denote the real and imaginary parts of $$f(z)$$ respectively. Then since we know that $$f(z)$$ consists of only the imaginary part, $$u(x,y) = 0$$, hence $$u_x=0, u_y=0$$. The Cauchy-Riemann Equations imply: $$\begin{cases} u_x=v_y\\u_y=-v_x\end{cases}$$

Hence, the Cauchy-Riemann equations can be represented by the following matrix. $$\begin {bmatrix} u_x &u_y\\-v_x & v_y \end {bmatrix}= \begin {bmatrix} 0 &0\\0 & 0\end {bmatrix}$$

So I have a few concerns at this point in the problem.

1. First, since the Jacobian matrix equals the $$\vec{0}$$, does that not imply that $$(0,0)$$ is the only place where this function is differentiable, hence contradicting the fact that $$f$$ is defined to be holomorphic (since there would be no other points in the neighborhood of $$\Omega$$)?
2. Second, if the Jacobian matrix is in fact the $$\vec{0}$$, then does that mean $$v(x,y)=v_0=0$$?

I would apprecaite some assistance on this problem. Thank you!

Any function $$v$$ with both partial derivative $$0$$ is a constant function. Since $$v_x$$ and $$v_y$$ are both $$0$$ it follows that $$v$$ is a constant. It appears that you are just over-thinking.