Must every Cauchy-Riemann condition be fulfilled simultaneously?

Question

Working through problems in my complex analysis book, and I have to determine where the derivative exists for a function. I know that the derivative can exist only along a certain curve, however I don't know if that must be true simultaneously or not. For instance, one such function I've been given is $f(z) = \frac{ix + 1}{y}$ which gives $u(x,y) = \frac{1}{y}$ and $v(x,y) = \frac{x}{y}$, and for the derivatives, you get $u_x = 0, v_y = \frac{-x}{y^2}, u_y = \frac{-1}{y^2}, and -v_x = \frac{-1}{y}$.

I know that this implies the derivative exists when $\frac{-x}{y^2} = 0$ (i.e. $x = 0$) and when $\frac{-1}{y^2} = \frac{-1}{y}$ (i.e. $y = 1$), but I don't know if this implies the derivative exists when EITHER of those things is true (derivative exists at $(0, a)$ and $(b, i)$) or only when BOTH of those things are true (derivative exists at $(0, i)$). I'm only asking for clarity on that front. Thank you!

Answer 1

For posterity: Although OP's question is answered in the comments, in general care is required to address these issues.

Let's agree that $z_{0}$ is a complex number, $u$ and $v$ are real-valued functions of a complex variable defined in some neighborhood of $z_{0}$ and having partial derivatives in this neighborhood, and that $f = u + iv$. There are at least four sets of criteria in play:

1. One Cauchy-Riemann equation, $u_{x} = v_{v}$ or $u_{y} = -v_{x}$, holds at $z_{0}$.
2. Both Cauchy-Riemann equations hold at $z_{0}$.
3. $f$ is complex-differentiable at $z_{0}$.
4. $f$ is holomorphic at $z_{0}$, i.e., complex-differentiable at each point of some neighborhood of $z_{0}$.

(Regarding 4., books and instructors often say holomorphic in a neighborhood of $z_{0}$ rather than holomorphic at $z_{0}$. But arguably either is correct, and mathematicians generally prefer minimal-sounding hypotheses.)

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As addressed in the comments, 3. implies 2., so 2. is necessary for complex-differentiability at $z_{0}$.

As for sufficiency:

• If the partial derivatives of $u$ and $v$ are continuous in a neighborhood of $z_{0}$, then the Cauchy-Riemann equations at $z_{0}$ imply complex differentiability at $z_{0}$. One proof is to use real multivariable calculus: Continuity of the partial derivatives in a neighborhood of $z_{0}$ implies $u$ and $v$ are real-differentiable at $z_{0}$. The C-R equations then express that $df(z_{0})$ is complex-linear, so $f$ is complex-differentiable at $z_{0}$. (The function in question here has continuous first partial derivatives where the partials are defined, so it falls under this bullet point.)
• It is possible for $f$ to be complex-differentiable at $z_{0}$ even if the partials of $u$ and/or $v$ are discontinuous at $z_{0}$. If we write $z = x + iy$ with $x$ and $y$ real, the function $$ f(z) = z\bar{z} \sin \frac{1}{z\bar{z}} = (x^{2} + y^{2}) \sin\frac{1}{x^{2} + y^{2}} $$ is complex differentiable at $0$ because $|f(z)| \leq |z|^{2}$, but its partials are discontinuous at $0$ (in fact, unbounded in every neighborhood of $0$).
• Existence of the partials in a neighborhood of $z_{0}$ does not even imply continuity of $u$ or $v$, so generally 2. does not imply 3. For example, $$ f(z) = \frac{\operatorname{im}(z^{2})}{z\bar{z}} = \frac{2xy}{x^{2} + y^{2}} $$ (and $f(0) = 0$) has partials equal to $0$ at $0$, but is discontinuous at $0$, hence not even real-differentiable at $0$.

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Finally (although not asked), the relationship between 3. and 4. may be in the minds of future readers. By definition, $f$ is holomorphic at $z_{0}$ if and only if $f$ is complex differentiable in some neighborhood of $z_{0}$. The italicized fine print is essential. For example, $$ f(z) = z\bar{z} = x^{2} + y^{2} $$ is complex differentiable at $0$, real-differentiable everywhere (as a real polynomial), but not complex-differentiable anywhere except $0$, and hence according to our definition holomorphic nowhere, not even at $0$.