# Can a function be analytic if it does not satisfy the Cauchy-Riemann conditions

I'm a student of electrical engineering, preparing for the theoretical exam covering complex analysis. I'm confused about the analyticity of a complex function, so I'm asking for clarification because our professor inadequately explained this concept.

What I understand from my lecture notes: suppose there's a complex-valued function $f(z)$. Such function is analytic if the following conditions are satisfied:

1. $f$ is differentiable at $z_0$, therefore $f'(z_0)$ exists
2. $f$ is differential at every point of some $\epsilon$-spherical neighborhood of $z_0$
3. $f$ can be expanded as a Taylor series in the vicinity of $z_0$

At that point, Cauchy-Riemann equations are nowhere mentioned. However, I've found online that C-R equations are a necessary condition for a complex-valued function being holomorphic (or analytic, although terms are used interchangeably). And here's where I'm starting to get lost.

If C-R equations are a necessary condition for holomorphicity, but they are not sufficient conditions for complex differentiability, then how can they ensure given function $f$ being holomorphic, if according to what I've read, such function must be differentiable at given point $z_0$ in order to be considered holomorphic? Can, therefore, a function be holomorphic (or analytic) if C-R conditions are not satisfied?

What I've read thus far

• C-R equations are necessary conditions for a function to be analytic. – Kavi Rama Murthy Jun 18 '18 at 10:03
• Alright, that's a start. So why is a complex differentiability mentioned? If a function must be smooth to be analytic (correct me if I'm wrong) and if C-R equations must be satisfied, what happens in a case when only first or second order derivatives exist? Is a function not analytic then despite satisfying C-R equations? – Pero Alex Jun 18 '18 at 10:09
• One of many interesting facts in Complex Analysis is if $f$ has continuous first partial derivatives satisfying C-R equations then $f$ is automatically infinitely differentiable. – Kavi Rama Murthy Jun 18 '18 at 10:18
• Well, that just made the whole thing clear for me. I hope I'm not asking too much, but can you provide a link to the proof for this claim? I'd be extremely helpful. Anyways, thank you for helping me understand this problem! Will mark the question solved. – Pero Alex Jun 18 '18 at 10:22
• Standard texts on CA are Conway and Rudin. Both these books have proofs of all the statements I have made. – Kavi Rama Murthy Jun 18 '18 at 10:28

## 1 Answer

this is a little late, and so I’m sure you have come across an answer by now! But I shall write the below nonetheless, as it may help others. Firstly we need to realize what the Cauchy-Riemann equations tell us. If you derive them (you can look this derivation up it comes from the definition of the derivative actually), you will see that they come from the assumption that a complex function is differentiable at the point of interest. So, if we are differentiable, then we satisfy Cauchy-Riemann. Which also tells us that the contrapositive is true, so that the statement “if we do not satisfy Cauchy-Riemann, then we are not differentiable” is true. Now onto analyticity, for a function to be analytic at the point P in the complex plane, it must be differentiable in a neighborhood of P. So Cauchy-Riemann must be satisfied in that neighborhood of P. It is a stricter condition, as we can be differentiable at a point but not necessarily analytic/holomorphic if we are not differentiable in the neighborhood.

So now let’s analyze your question. You have asked “can a function be analytic if it doesn’t satisfy CR?” Let’s do a proof of falsity by contradiction: assume a function can be analytic and not satisfy CR at the point P (I’m making your question slightly more specific). This would mean that we are differentiable in a neighborhood of a point P but do not satisfy CR at the point P. If we do not satisfy CR at the point P, then we are not differentiable at P (as per the contrapositive statement above). But that would mean that we are not differentiable in a neighborhood of P since the neighborhood of P includes itself! So we arrive at our contradiction, that we started as differentiable and became non-differentiable in a neighborhood of P which contradict one another, so we abandon the assumption that we are analytic. Thus, if we do not satisfy CR at a point P, then we cannot be analytic at the point P.

Now suppose that your question became “can we be analytic if we do not satisfy Cauchy-Riemann at some point or set of points in the neighborhood of P?” Observe that this is as general as we can make your question! The proof to this is just the same as above, assume you can be analytic at P and do not satisfy CR at some point or set of points in a neighborhood of P, then we are not differentiable at some point or set of points in a neighborhood of P, so we are not differentiable in the neighborhood of P, and so we are lead to the same contradiction as above. To be specific, it is the contradiction that we started as differentiable but had information (CR not satisfied) to show that we are not differentiable and so we are lead to the contradiction and therefore must abandon the assumption.

So we can never be analytic at a point P without satisfying the Cauchy-Riemann equations in a neighborhood of P. This is the ultimate punch line of the demonstration above. I hope things have been made more clear from this, it is good practice in mathematics to try and prove claims that you make or potential answers to questions you have, it allows you to practice what you’ve learned as well and maybe even reinforce your knowledge or learn something new!

Have a great day.

• I don't know why your answer was downvoted, because it was actually well written. Although I have passed my math exam a while time ago (2 days after posting this question), complex analysis is still perplexing me. Because of sites like this, students like myself can progress faster. And it's all thanks to people like you that decide to invest their time in answering questions. Although I've gotten my answer in comments (as you can see), I respect you coming here and replying. Thank you! – Pero Alex Oct 27 '18 at 7:22
• Hey Pero, thanks for the reply! Yes I noticed after posting that you had already been helped, I just zeroed in on the fact that the question claimed to have “zero answers” and so without looking at the other comments just posted one as we had recently covered similar material in one of my complex courses! I hope it helped, and I think that everyone gets perplexed from time to time which is why sites like these are so beneficial! Have a great day. – Amun Patel Oct 29 '18 at 1:16