Complex Differentiability and the Cauchy–Riemann Equations: Examples and Counterexamples

For every nonzero increment $h$, $\frac{f(z+h)-f(z)}{h}=\frac{(z+h{)}^2-z^2}{h}=2z+h\mathrm{.}$

Since $\mathrm{\mid }(2z+h)-2z\mathrm{\mid }=\mathrm{\mid }h\mathrm{\mid }\to 0$, the quotient in step 1.1 tends to $2z$. Thus $f^{\mathrm{\prime }}(z)=2z$ by [L1].

The point $z$ was arbitrary, so $f$ is entire.

Writing $z=x+iy$ gives $z^2=(x^2-y^2)+i(2xy)$. Hence $u_x=2x=v_y$ and $u_y=-2y=-v_x$, in agreement with step 2.1.

Moreover $u_{xx}+u_{yy}=2-2=0$ and $v_{xx}+v_{yy}=0+0=0$. The polynomial components are $C^2$, so [L3] identifies them as harmonic and identifies $v$ as a harmonic conjugate of $u$.

For $h\mathrm{\ne }0$, the addition law and defining series give $\frac{e^{z+h}-e^z}{h}=e^z\frac{e^h-1}{h}=e^z(1+{\sum }_{n\ge 2}\frac{h^{n-1}}{n!})\mathrm{.}$

From [L3], $u=e^x\cos y$ and $v=e^x\sin y$. Therefore $u_x=e^x\cos y=v_y$ and $u_y=-e^x\sin y=-v_x$, and $u_x+i{v}_x=e^{x+iy}$.

Put $c:=\cos \theta$, $s:=\sin \theta$, $A:=e^{rc}$, and $\varphi :=rs$. Then $U=A\cos \varphi$ and $V=A\sin \varphi$, so $U_r=A(c\cos \varphi -s\sin \varphi ),V_r=A(c\sin \varphi +s\cos \varphi ),$ $U_{\theta }=-r{V}_r,V_{\theta }=r{U}_r\mathrm{.}$

When $0<\mathrm{\mid }h\mathrm{\mid }\le 1$, absolute convergence at $1$ gives the finite constant $C:={\sum }_{n\ge 2}1\mathrm{/}n!$, and $\mid {\sum }_{n\ge 2}\frac{h^{n-1}}{n!}\mid \le \mathrm{\mid }h\mathrm{\mid }{\sum }_{n\ge 2}\frac{1}{n!}=C\mathrm{\mid }h\mathrm{\mid }⟶0.$

Thus the quotient tends to $e^z$, directly confirming ${\exp }^{\mathrm{\prime }}(z)=\exp z$ and [L4].

Since $r>0$, step 1.3 is equivalent to $U_r=r^{-1}{V}_{\theta }$ and $V_r=-r^{-1}{U}_{\theta }$, exactly [L5]. At $r=0$ these polar identities are not asserted; the Cartesian calculation in step 1.2 covers the origin.

If $0<\mathrm{\mid }h\mathrm{\mid }<\mathrm{\mid }z\mathrm{\mid }\mathrm{/}2$, then $\mathrm{\mid }z+h\mathrm{\mid }\ge \mathrm{\mid }z\mathrm{\mid }-\mathrm{\mid }h\mathrm{\mid }>\mathrm{\mid }z\mathrm{\mid }\mathrm{/}2>0$, so $z+h$ remains in the punctured plane.

For such $h$, $\frac{f(z+h)-f(z)}{h}=-\frac{1}{z(z+h)}\mathrm{.}$

The error from the proposed derivative satisfies $\mid -\frac{1}{z(z+h)}+\frac{1}{z^2}\mid =\frac{\mathrm{\mid }h\mathrm{\mid }}{\mathrm{\mid }z{\mathrm{\mid }}^2\mathrm{\mid }z+h\mathrm{\mid }}\le \frac{2\mathrm{\mid }h\mathrm{\mid }}{\mathrm{\mid }z{\mathrm{\mid }}^3}⟶0.$

Hence $f^{\mathrm{\prime }}(z)=-1\mathrm{/}{z}^2$ by [L1]. Since $z$ was arbitrary in $\mathbb{C}\setminus \{0\}$, $f$ is holomorphic there. The point $0$ is absent from the function's domain, not merely exceptional for the derivative formula.

On the set $D:=\{z:cz+d\mathrm{\ne }0\}$, [L1] gives $F^{\mathrm{\prime }}(z)=\frac{a(cz+d)-c(az+b)}{(cz+d{)}^2}=\frac{ad-bc}{(cz+d{)}^2}\mathrm{.}$

Both numerator and denominator in step 1.1 are nonzero on $D$, so $F^{\mathrm{\prime }}(z)\mathrm{\ne }0$ throughout $D$.

Thus [L2] makes $F$ orientation-preserving conformal at every point of $D$.

If $c=0$, then $ad\mathrm{\ne }0$, hence $d\mathrm{\ne }0$ and $D=\mathbb{C}$. If $c\mathrm{\ne }0$, the excluded equation $cz+d=0$ has the unique solution $z=-d\mathrm{/}c$, so $D=\mathbb{C}\setminus \{-d\mathrm{/}c\}$. No global injectivity claim on $\mathbb{C}$ is needed.

Expanding gives $u=x^2-y^2$ and $v=2xy$. On $x=c$, $(u,v)=(c^2-y^2,2cy)$.

At a crossing $z=c+id\mathrm{\ne }0$, [L1] gives $f^{\mathrm{\prime }}(z)=2z\mathrm{\ne }0$. Hence [L2] says the real derivative is an oriented similarity, so it carries the perpendicular vertical and horizontal tangent directions to perpendicular nonzero tangent directions of the two image curves.

If $c\mathrm{\ne }0$, eliminating $y=v\mathrm{/}(2c)$ gives $u=c^2-v^2\mathrm{/}(4c^2)$, a left-opening parabola. If $c=0$, the image is $(u,v)=(-y^2,0)$, the nonpositive real ray, which is the degenerate member of that family.

On $y=d$, $(u,v)=(x^2-d^2,2dx)$. If $d\mathrm{\ne }0$, eliminating $x=v\mathrm{/}(2d)$ gives $u=v^2\mathrm{/}(4d^2)-d^2$, a right-opening parabola. If $d=0$, the image is $(u,v)=(x^2,0)$, the nonnegative real ray.

At $z=0$, the derivative is zero, so [L2] gives no conformality conclusion; indeed the two degenerate rays meet there. Steps 2.1 and 2.2 cover zero and nonzero grid parameters, and step 1.2 covers exactly the noncritical crossings.

By [L1], $f$ is the real-linear map $(x,y)\mapsto (x,-y)$. Its increment is exactly its linear action, so it is real totally differentiable everywhere with derivative matrix $\mathrm{diag}(1,-1)$.

Directly, for nonzero real $t$ the quotient $(\overline{z+t}-\bar{z})\mathrm{/}t$ is $1$, whereas the quotient for the increment $it$ is $(\overline{z+it}-\bar{z})\mathrm{/}(it)=-1$. Thus the complex difference quotient has incompatible directional limits.

Its components $u=x$ and $v=-y$ have $u_x=1$ and $v_y=-1$, so the first Cauchy–Riemann equation fails at every point. By [L2], $f$ is nowhere complex differentiable.

The same map is real totally differentiable everywhere by step 1.1 and complex differentiable nowhere by steps 1.2 and 2.1, so it refutes the claim.

At $z=0$ and $h\mathrm{\ne }0$, $\frac{f(h)-f(0)}{h}=\frac{\mathrm{\mid }h{\mathrm{\mid }}^2}{h}=\bar{h},$ whose modulus is $\mathrm{\mid }h\mathrm{\mid }$ and hence tends to $0$. Thus $f^{\mathrm{\prime }}(0)=0$.

At $z=x+iy$, the components are $u=x^2+y^2$ and $v=0$, so $u_x=2x$, $u_y=2y$, and $v_x=v_y=0$.

If $f$ were complex differentiable at a nonzero $z$, [L2] would force $2x=0$ and $2y=0$, hence $z=0$, a contradiction. Therefore $f$ is not complex differentiable at any nonzero point.

Steps 1.1 and 2.1 cover every $z\in \mathbb{C}$, so the complex-differentiability locus is exactly $\{0\}$. Every open neighbourhood of $0$ contains a nonzero point, where step 2.1 gives failure; thus [L1] says $f$ is holomorphic on no neighbourhood of $0$.

In real coordinates, $f(x+iy)=x+i0$, so it is the real-linear map $(x,y)\mapsto (x,0)$ and is real totally differentiable everywhere with that same linear map as derivative.

Its components have $u_x=1$ and $v_y=0$. The first Cauchy–Riemann equation therefore fails everywhere, so [L1] makes $f$ nowhere complex differentiable.

Equivalently, at any $z$, the difference quotient is $1$ along nonzero real increments and $0$ along nonzero imaginary increments. These incompatible limits independently confirm step 2.1 and refute the claim.

At $0$, for nonzero real $t$ the difference quotient is $\mathrm{\mid }t\mathrm{\mid }\mathrm{/}t$, which equals $1$ for $t>0$ and $-1$ for $t<0$. Thus no complex derivative exists at $0$.

Let $z=x+iy\mathrm{\ne }0$ and put $r=\mathrm{\mid }z\mathrm{\mid }>0$. Rationalizing the real-coordinate increments gives $u_x={\lim }_{t\to 0}\frac{\sqrt{(x+t{)}^2+y^2}-r}{t}={\lim }_{t\to 0}\frac{2x+t}{\sqrt{(x+t{)}^2+y^2}+r}=\frac{x}{r},$ and similarly $u_y=y\mathrm{/}r$, while the imaginary component $v$ is identically zero.

If $f$ were complex differentiable at this nonzero point, [L1] would force $u_x=v_y=0$ and $u_y=-v_x=0$. Step 1.2 would then give $x=y=0$, contradicting $z\mathrm{\ne }0$.

Step 1.1 covers the origin and step 2.1 covers every nonzero point. Hence the modulus map is nowhere complex differentiable.

The polynomial partials are continuous everywhere and satisfy $u_x=3x^2+3y^2=v_y,u_y=6xy=v_x\mathrm{.}$

The second Cauchy–Riemann equation is $u_y=-v_x$, so by step 1.1 it holds exactly when $6xy=-6xy$, equivalently $xy=0$.

At every point with $xy=0$, [L1] and steps 1.1–2.1 give complex differentiability. At every point with $xy\mathrm{\ne }0$, [L2] and step 2.1 rule it out. Thus the differentiability locus is exactly the union of the coordinate axes.

Every open ball about any point of either axis contains a point with both coordinates nonzero: for radius $r>0$, a sufficiently small displacement in both coordinate directions supplies one. Every open ball about a point off the axes already contains its centre, where differentiability fails. Hence no nonempty open set consists entirely of differentiability points.

Holomorphy at a point requires complex differentiability throughout some open neighbourhood there. Step 4.1 therefore shows that $f$ is holomorphic nowhere, despite being complex differentiable at every point of both axes.

For $z\mathrm{\ne }0$, $\mathrm{\mid }f(z)\mathrm{\mid }=\mathrm{\mid }\bar{z}{\mathrm{\mid }}^2\mathrm{/}\mathrm{\mid }z\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$. Hence $f(z)\to 0=f(0)$ as $z\to 0$, so the example is even continuous at the point in question.

On the real axis, $f(t)=t$; on the imaginary axis, $f(it)=it$. Therefore at the origin $u_x=1,v_x=0,u_y=0,v_y=1.$

For $z\mathrm{\ne }0$, the derivative quotient at the origin is $\frac{f(z)-f(0)}{z}=\frac{{\bar{z}}^2}{z^2}\mathrm{.}$

Thus $u_x=v_y$ and $u_y=-v_x$ at $0$: both Cauchy–Riemann equations hold there.

Along nonzero real $z=t$, the quotient in step 1.3 is $1$. Along $z=t(1+i)$, it is $(1-i{)}^2\mathrm{/}(1+i{)}^2=-1$. Both paths tend to $0$, so [L1] shows that $f^{\mathrm{\prime }}(0)$ does not exist.

Steps 1.2 and 2.1 verify the false claim's entire hypothesis at $0$, while step 2.2 denies its conclusion. This also exhibits the missing ingredient in [L2]: the coordinate map is not real totally differentiable at $0$.

On $\mathbb{C}\setminus \{0\}$, the power, reciprocal, exponential, and chain rules show that $f$ is holomorphic. Consequently all four partials exist and satisfy the Cauchy–Riemann equations there by [L1], [L2], and [L4].

For nonzero real $t$, both $t^4$ and $(it{)}^4$ equal $t^4$, so $f(t)=f(it)=e^{-1\mathrm{/}{t}^4}\in \mathbb{R}\mathrm{.}$

Put $x=1\mathrm{/}\mathrm{\mid }t\mathrm{\mid }$. As $t\to 0$, $x\to +\mathrm{\infty }$; for $x\ge 1$, $0\le x{e}^{-x^4}\le x{e}^{-x}$, and [L3] with $m=1$, $a=1$ makes the last expression tend to $0$. Hence $e^{-1\mathrm{/}{t}^4}\mathrm{/}\mathrm{\mid }t\mathrm{\mid }\to 0$.

Along $z=t(1+i)$ with nonzero real $t$, one has $z^4=-4t^4$, and therefore $f(t(1+i))=e^{1\mathrm{/}(4t^4)}⟶+\mathrm{\infty }(t\to 0)\mathrm{.}$ So $f$ is unbounded in every neighbourhood of $0$ and is not continuous there.

Steps 1.2–1.3 give $u_x(0)=u_y(0)=v_x(0)=v_y(0)=0$. Thus all four partials exist at $0$ and satisfy both Cauchy–Riemann equations there. Together with step 1.1, the false claim's hypotheses hold throughout the open set $\mathbb{C}$.

By [L4], the discontinuity in step 1.4 rules out complex differentiability at $0$. Hence $f$ satisfies Cauchy–Riemann everywhere but is not holomorphic on $\mathbb{C}$, refuting the claim.

There is also a Wirtinger warning. Off $0$, $f$ is holomorphic, so conjugating its Cauchy–Riemann equations gives $(\bar{f}{)}_z=0$; at $0$, step 2.1 gives the same value. Thus $(\bar{f}{)}_z$ is identically zero and continuous. Nevertheless, off $0$ the chain rule gives $f^{\mathrm{\prime }}(z)=4z^{-5}{e}^{-z^{-4}}$, whose modulus along $z=t(1+i)$ tends to infinity, so the coordinate partials of $f$ and $\bar{f}$ are not continuous at $0$.

Let $z=x+iy\in U$ and choose $r:=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2>0$. If $\mathrm{\mid }w-z\mathrm{\mid }<r$, then $\mathrm{\mid }\mathrm{Re}w-x\mathrm{\mid }\le \mathrm{\mid }w-z\mathrm{\mid }<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$, so $\mathrm{Re}w$ has the same sign as $x$. Thus $B(z,r)\subseteq U$, and [L1] shows that $U$ is open.

But $-1,1\in U$ and $f(-1)=0\mathrm{\ne }1=f(1)$, so $f$ is not constant on $U$.

The same ball $B(z,r)$ lies wholly in one half-plane, so $f$ is constant on it. For every sufficiently small nonzero $h$ with $z+h\in U$, the difference quotient $(f(z+h)-f(z))\mathrm{/}h$ is therefore $0$. Hence $f^{\mathrm{\prime }}(z)=0$.

Since $z\in U$ was arbitrary, [L2] says $f$ is holomorphic on $U$ and has derivative zero everywhere there.

Steps 3.1 and 1.2 refute the claim. The missing hypothesis is connectedness: $U$ is the disjoint union of two nonempty open half-planes, whereas [L3] requires a domain to be connected.