Question About Differentiability Of Principal Branch Of log(z).

Question

Theorem:1 Let the function $$f (z) = u(r, θ) + iv(r, θ)$$be defined throughout some neighborhood of a nonzero point $z_0 = r_0 \exp(iθ_0)$ and suppose that

(a) The first-order partial derivatives of the functions u and v with respect to r and θ exist everywhere in the neighborhood;

(b) Those partial derivatives are continuous at $(r_0, θ_0)$ and satisfy the polar form $$ru_r = v_θ , u_θ = −rv_r$$of the Cauchy–Riemann equations at $(r_0, θ_0)$.

Then $f$ is differentiable at $z_0$.

Now consider

$$\text{Log}(r\exp(i\theta))=\ln r+i\theta$$

for $(r, θ)\in \mathbb R^+\times (-\pi,\pi].$

Then

$u_{\theta}=v_r=0$ and $u_r=\frac 1 r$ ,$v_{\theta}=1$.

So clearly all Hypothesis of Theorem 1 satisfy for$(r, θ)\in \mathbb R^+\times (-\pi,\pi].$.so Log(z) is differentiable everywhere except origin . which is false since it is not even continuous on negative real axis.

So what I am missing here ?

Edit:

Given $$f(z) = u(r, \theta) + iv(r, \theta)$$

for $(r,\theta)\in\underbrace{(0,\infty) \times (a,a+2\pi]}_{D_a}$. $z_0 = r_0 \exp(i\theta_0)$ , $(r_0,\theta_0)\in D_a$. for some $a\in \mathbb R$.

and Map $$p_a : D_a\to \mathbb C^*$$

Now we can use theorem 1 for $z_0$ if there exist $V\subset D_a$$\underbrace{\text{open}}_{Question:1}$ containing $(r_0,\theta_0)$. and there exist $z_0 \in U\subset \mathbb C^*$ open such that restriction

$$p_a : V\to U $$ $\underbrace{\text{is Homeomorphism.}}_{Question:2}$

Assuming this is true for given $z_0$ now we have

$$f : U \xrightarrow{p_a^{-1}} V \xrightarrow{\phi} \mathbb C $$with $\phi = u + iv$.

Now $f$ is differentiable at $z_0=r_0\cos(\theta_0)+ir_0\sin(\theta_0)~$in complex sense If and only if it is differentiable at $(r_0\cos(\theta_0),r_0\sin(\theta_0))~$as function from subset of $\mathbb R^2$ and satisfy Cauchy Reimann equation at $(r_0\cos(\theta_0),r_0\sin(\theta_0))~$.

By hypothesis of theorem 1 ,$\phi$ differentiable at $(r_0,\theta_0)$. and since $p_a$ is of class $C^1$ with non zero Jacobian we have by inverse function theorem $p_a^{-1}$ is differentiable at $(r_0\cos(\theta_0),r_0\sin(\theta_0))~$.Hence by chain rule $f$ is differentiable at $(r_0\cos(\theta_0),r_0\sin(\theta_0))~$.

Moreover by using matrix for derivative and polar form for Cauchy Reimann equation we have Cauchy Reimann equation for $f$ at $z_0$. So eventually we have $f$ differentiable In complex sense at $z_0$.

So the Point I missed is that points on negative real axis are not interior point of $(0,\infty)\times (-\pi,\pi]$.And to use chain rule it is essential.

Question 1: There by open means we need $V$ is open in $D_a$ as a subspace topology or we need just $(r_0,\theta_0)$ to be interior point.

Question 2: Where we use the fact that restriction is homeomorphism.

Question 3: Here I restricted domain of $f$ to be $D_a$ . But if domain was $(0,\infty)\times \mathbb R$. Then we don't need to check "some neighborhood of a nonzero point $z_0$". right? .

Answer 1

Working with polar coordinates means that we use the function $$p : (0,\infty) \times \mathbb R \to \mathbb C^* = \mathbb C \setminus \{0\}, p(r,\theta) = r\exp(i \theta) = r\cos \theta + i r\sin \theta$$ to parameterize the points of $\mathbb C^*$. This function surjective, but not injective. Actually we have $p(r,\theta) = p(r',\theta')$ iff $r' = r$ and $\theta' = \theta + 2k\pi$ for some $k \in \mathbb Z$.

$p$ is continuously differentiable in the sense of multivariable calculus and has non-vanishing Jacobian determinant. Thus it is an open map.

The restrictions $$p_a : (0,\infty) \times (a,a+2\pi] \to \mathbb C^*$$ $$\tilde p_a : (0,\infty) \times [a,a+2\pi) \to \mathbb C^*$$ are bijective for all $a \in \mathbb R$, but they are no homeomorphisms. In fact $p_a^{-1}$ and $\tilde p_a^{-1}$ are not continuous at the points of the ray $R_a = \{re^{ia} \mid r \in (0,\infty)\}$. However, since $p$ is an open map, $$\bar p_a : (0,\infty) \times (a,a+2\pi) \to \mathbb C^*$$ is always a homeomorphism onto its image $\mathbb C_a = \mathbb C^* \setminus R_a$ (which is a sliced plane).

For a correct use Theorem 1 we need to understand the meaning of

Let the function $$f(z) = u(r, \theta) + iv(r, \theta)$$ be defined throughout some neighborhood of a nonzero point $z_0 = r_0 \exp(i\theta_0)$.

Writing $f(z) = u(r, \theta) + iv(r, \theta)$ on some neighborhood of $z_0$ means more precisely that we have

1. an open neighborhood $U \subset \mathbb C^*$ of $z_0$,

2. a parameterization of $U$ by $p$, i.e. a subset $V \subset (0,\infty) \times \mathbb R$ which is mapped by $p$ bijectively onto $U$

so that $f$ has the form

$$f : U \xrightarrow{p_V^{-1}} V \xrightarrow{\phi} \mathbb C $$ with $\phi = u + iv$. Here $p_V : V \to U$ is the bijection obtained by restricting $p$.

The proof of Theorem 1 relies on the chain rule. To apply the chain rule, we need to require that

• $p_V^{-1}$ is differentiable at $z_0$.
  Here we regard $p_V^{-1}$ as a map $p_V^{-1} : U \to (0,\infty) \times \mathbb R$. Differentiability is automatically guaranteed if $V$ is open in $(0,\infty) \times \mathbb R$. This follows from the inverse function theorem (recall that $p$ is continuously differentiable and has non-vanishing Jacobian determinant). By no means we must take $U = \mathbb C^*$ and parameterize it by $p_a$ or $\tilde p_a$ for $z_0 \in R_a$. The inverses of these functions are not continuous (hence not differentiable) at the points of $R_a$.

• $V$, the domain of $\phi$, is open in $(0,\infty) \times \mathbb R$ and $\phi$ is differentiable at $(r_0,\theta_0)$.
  More generally, it suffices to know that $(r_0,\theta_0)$ is an interior point of $V$. In this case there exists an open neighborhood $V'$ of $(r_0,\theta_0)$ such that $V' \subset V$ and we may replace $U$ by $U' = p(V')$ which is open in $\mathbb C^*$ since $p$ is an open map.
  In other words, if $(r_0,\theta_0)$ is an interior point of $V$, we may assume w.l.o.g. that $V$ itself is open.

The second issue (openness of $V$) is frequently not an essential problem. Indeed, even if $V$ is not open, $\phi$ can be often be defined on an open $W \subset (0,\infty) \times \mathbb R$ such that $V \subset W$. In this case we can regard $p_V^{-1}$ as a map $p_V^{-1} : U \to W$ and write $$f : U \xrightarrow{p_V^{-1}} W \xrightarrow{\phi} \mathbb C .$$

For example, for the $\operatorname {Log}$-function we can take $\phi_{\operatorname {Log}}(r,\theta) = r\exp{i\theta}$ which is defined on all of $(0,\infty) \times \mathbb R$ and is differentiable at all points.

Summarizing, to solve both issues, we must require that $V$ is open. That is, we require that we are given an open subset $V \subset (0,\infty) \times \mathbb R$ which is mapped by $p$ bijectively onto $U$. In that case $p_V : V \to U$ is a homeomorphism (because $p$ is an open map). Thus we strengthen 2. to

• a homeomorphic parameterization of $U$ by $p$, i.e. an open subset $V \subset (0,\infty) \times \mathbb R$ which is mapped by $p$ homeomorphically onto $U$.

For an open $V$ we know that $p_V^{-1} : U \to V$ is differentable and its Jacobian matrix at $z_0$ is $$Jp_V^{-1}(z_0) = (Jp_V(r_0, \theta_0))^{-1} .$$ By assumption $\phi$ satisfies the Cauchy–Riemann equations in polar form. By the chain rule the Jacobian of $f$ at $z_0$ is

$$Jf(z_0) = J\phi(r_0, \theta_0) \cdot Jp_V^{-1}(z_0)$$ which allows to derive the Cauchy–Riemann equations for $f$.

Let us finally conisder the $\operatorname {Log}$-functon in your question. As explained above, we must not work with the merely bijective parameterization $p_{-\pi}$ of $\mathbb C^*$. We can of course write $$\operatorname {Log} : \mathbb C^* \xrightarrow{p_{-\pi}} (0,\infty) \times \mathbb R \xrightarrow{\phi_{\operatorname {Log}}} \mathbb C $$

The problem is that $p_{-\pi}^{-1}$ is not continuous at the points of the negative $x$-axis. Thus $\operatorname {Log}$ is not differentiable at these points - it is not even continuous. Theorem 1 does not apply at these points.

Answer 2

The point is that in Theorem 1, you must choose $\theta$ close to $\theta_0$ and $\theta$ is not unique. In your definition of the log, you choose $\theta \in (-\pi,\pi]$. If you take for example $z_0 = -1$ (or any other negative number), you have $\theta_0 = \pi$ and for all $\varepsilon > 0$ close to $0$, the argument of $-1 - \varepsilon i$ is close to $-\pi$. Therefore, the argument is not continuous hence not differentiable.

One way to solve the problem is to exclude the negative numbers. In this case, you can choose $\theta$ in $(-\pi,\pi)$ and the argument function becomes continuous (and differentiable). In Theorem 1, it is assumed (but not explicitely said) that you always choose $\theta$ close to $\theta_0$. If you take $\theta \in [0,2\pi)$ instead, you have a log which is differentiable at the negative numbers for not continuous at the positive numbers. If you take it in $(-\pi/2,3\pi/2]$, you have to exclude the pure imaginary numbers with negative imaginary part.