# Is it true that if function $f$ is analytic( ~ holomorphic) in $\Omega \subset\Bbb C$, then it satisfies C-R equations?

Is it true that if function $f$ is analytic ($\sim$ holomorphic) in $\Omega \subset \Bbb C$, then it satisfies the Cauchy-Riemann (C-R) equations? And is it t true that if $f$ satisfies C-R equations and the functions $u(x,y)$ and $v(x,y)$ (the real and imaginary parts of $f(x,y)$ respectively) have first partial derivatives which are continuous, then $f$ is analytic ($\sim$ holomorphic)?

• Yes, it's true. – Ahriman Aug 24 '12 at 16:19
• The answers are "yes". I am not sure you have a clear picture of the situation. What do you mean by "analytic" and "holomorphic"? They turn out to be equivalent definitions, but in the beginning they may differ. – Siminore Aug 24 '12 at 16:42
• See this answer, given a few hours ago: math.stackexchange.com/questions/186227/… – Christian Blatter Aug 24 '12 at 17:53
• Note that the Looman - Menchoff Theorem gives a stronger converse to the Cauchy - Riemann Equations, without requiring $f$ to be $C^1$. – EuYu Aug 24 '12 at 18:21
• Why do I feel that this is a copy-paste from the question linked to by Christian? – timur Aug 24 '12 at 18:37

Analytic and holomorphic are a priori different. A complex-analytic function is one that can be written as series with terms of the form $a_nz^n$. A holomorphic function is one that satisfies the Cauchy-Riemann equations. Your question is if they are equivalent, and the answer is yes.
Analytic functions are holomorphic, because you can differentiate the series term by term, and each term satisfies the Cauchy-Riemann equations. It is important here to distinguish complex-analyticity from real-analyticity, because for real-analytic functions this result does not hold. A real-analytic function in 2D is one that can be written as series with terms of the form $a_{n,m}x^ny^m$.