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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!

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1 Answer 1

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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.

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