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Let $f(z) = \ln z := \ln |z| + \arg (z)i$. Then the derivative is (if it exists) by definition:
$$\lim_{h\to 0}\frac{\ln (z+h)-\ln (z)}{h}=\lim_{h\to 0}\frac{\ln |z+h| +\arg(z+h)i-\ln |z| -\arg(z)i }{h}$$

AlvinL
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  • Have you seen the Cauchy-Riemann equations? – StackTD Feb 23 '16 at 20:24
  • what's $u(x,y), v(x,y)$ here? – AlvinL Feb 23 '16 at 20:25
  • You wrote $\ln z = \ln |z| + i \arg z$; notice that $\ln |z|$ and $\arg z$ are real, so $u$ and $v$ are...? – StackTD Feb 23 '16 at 20:27
  • oh, $u =\ln |z|\cos \arg(z)$ and $v =\ln |z|\sin\arg z$ – AlvinL Feb 23 '16 at 20:27
  • Almost; in $i\arg z$, $\arg z$ is real so... – StackTD Feb 23 '16 at 20:28
  • ok, but that helps to solve the mystery whether $f$ is differentiable unless it also lets us compute the derivative? – AlvinL Feb 23 '16 at 20:30
  • Checking the Cauchy-Riemann equations will give you differentiability but you can also use it to find the derivative; I can elaborate in an answer if necessary? – StackTD Feb 23 '16 at 20:31
  • please do so, Please explain in general case, thanks. – AlvinL Feb 23 '16 at 20:32
  • the derivative of the principal branch of $\ln z$ on $\mathbb{C}$ is $\frac{d\ln (x+iy)}{dx} = \frac{1}{x+iy}$, $\frac{d\ln (x+iy)}{dy} = \frac{i}{x+iy} + 2 i \pi \delta(y) H(x)$ where $\delta$ is the Dirac delta and $H$ the Heaviside function, and $2 i \pi \delta(y) H(x)$ is the branch cut on $]0;\infty[$ – reuns Feb 23 '16 at 20:57

2 Answers2

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I'll use $\log$ for the complex logarithm and $\ln$ for the real-valued logarithm; you then have: $$\log z = \ln |z| + i \arg z = \ln r + i\varphi$$ where I use $r = |z|$ and $\varphi = \arg z$ for simplicity. In this form, we have written $$\log z = u(x,y)+iv(x,y)$$ with $u(x,y) = \ln r = \ln \sqrt{x^2+y^2} = \tfrac{1}{2}\ln(x^2+y^2)$ and $v(x,y) = \varphi$.

You can check the Cauchy-Riemann equations yourself and find (*) $$\frac{\partial u}{\partial x} = \frac{x}{x^2+y^2} = \frac{\partial \varphi}{\partial y} \quad ; \quad \frac{\partial u}{\partial y} = \frac{y}{x^2+y^2} = - \frac{\partial \varphi}{\partial x}$$ Use this to find the derivative directly: $$\frac{\mbox{d} \log z}{\mbox{d}z} = \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} = \frac{x-iy}{x^2+y^2} = \frac{1}{z}$$

Remark: depending on how you define the complex logarithm, there will be different ways to find its derivative.

(*) The derivatives for $u$ are easy. For $v$, we have:

$$\left\{ \begin{array}{ccc} x = r\cos\varphi \\ y = r\sin\varphi \end{array}\right.$$ where both $r$ and $\varphi$ are (implicit) functions of $x$ and $y$. Differentiating both equations w.r.t. $x$ gives: $$\left\{ \begin{array}{ccc} 1 = \cos\varphi\frac{\partial r}{\partial x}-r\sin\varphi\frac{\partial \varphi}{\partial x} \\ 0 = \sin\varphi\frac{\partial r}{\partial x}+r\cos\varphi\frac{\partial \varphi}{\partial x} \end{array}\right.$$ This is a linear system of two equations in the variables $\frac{\partial r}{\partial x}$ and $\frac{\partial \varphi}{\partial x}$; solve for $\frac{\partial \varphi}{\partial x}$. In the same way, take the derivative w.r.t. $y$ and solve for $\frac{\partial \varphi}{\partial y}$. Can you take it from here?

StackTD
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  • confused about one bit. Namely how do you take the partial derivative of $\varphi$? I can't see any expression that lets you say its partial with respect to $y$ is exactly that. – AlvinL Feb 23 '16 at 20:45
  • I didn't work that out; use $x = r\cos\varphi$ and $y = r\sin\varphi$; e.g. continue by taking the derivative (implicit differentiation) w.r.t. $x$ of both equations and solve the system you obtain for $\tfrac{\partial \varphi}{\partial x}$, etc. – StackTD Feb 23 '16 at 20:46
  • Alright, I will work on, it, thank you for your time. – AlvinL Feb 23 '16 at 20:47
  • Yes, I edited - that happens when you submit before checking what you wrote ;o). – StackTD Feb 23 '16 at 20:48
  • StackTD one problem arises, when we consider $\varphi =\arctan (y/x)$, depending on the quadrant we sometimes have to add or subtract $\pi$, would that affect anything$ – AlvinL Feb 23 '16 at 21:35
  • If you calculate the derivatives from the polar formulas above (with implicit differentiation), there's no need for an explicit formula for $\varphi$, so you can avoid the trouble with $\arctan$. – StackTD Feb 23 '16 at 21:42
  • Let us continue this discussion in chat. – AlvinL Feb 23 '16 at 21:44
  • You need to restrict the domain here inasmuch as the complex logarithm isn't even continuous in the plane. – Mark Viola Feb 23 '16 at 22:32
  • I'm assuming the usual domain is taken ($\mathbb{C} \setminus {x ;\vert ; x\le 0 }$). – StackTD Feb 24 '16 at 09:02
  • @StackTD Many thanks for taking the time, I've worked it out and understand now :) – AlvinL Feb 24 '16 at 10:21
  • The derivative, however, by my calculations is $f'(z) = u_x(x,y) + iv_x(x,y)$ – AlvinL Feb 24 '16 at 10:41
  • Correct, that was a typo! Fixed. – StackTD Feb 24 '16 at 10:43
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Let $ U =\mathbb C \setminus (-\infty,0].$ Then $f(z)=\ln |z| + i\arg z$ is continuous on $U$ and we have $e^{f(z)} = z$ there. This shows $f(z)$ is injective on $U.$ Fix $z\in U.$ Then for small nonzero $h$ we have

$$1=\frac{e^{f(z+h)}-e^{f(z)}}{h} = \frac{e^{f(z+h)}-e^{f(z)}}{f(z+h) - f(z)}\frac{f(z+h) - f(z)}{h}.$$

The injectivity of $f$ shows that $f(z+h) - f(z)\ne 0,$ so we're OK dividing by it above. As $h\to 0, f(z+h) \to f(z)$ by the continuity of $f.$ So the first difference quotient on the right tends to the derivative of $e^w$ at $w=e^{f(z)},$ which is $e^{f(z)} = z \ne 0.$ Knowing all of this, we can now write

$$\tag 1 \frac{f(z+h) - f(z)}{e^{f(z+h)}-e^{f(z)}} = \frac{f(z+h) - f(z)}{h}.$$

Since the left side of $(1) \to 1/z,$ we get $f'(z) = 1/z$ (as expected).

zhw.
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