I was reading Needham's Visual Complex Analysis, and could not figure out how we get length $Ld\theta$ here:

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It is just an approximation of the arc length of that sector with radius $L$ and angle $dθ$. Note that such a method as presented in that text is totally not rigorous, and will not help you learn to rigorously prove anything in complex analysis!

  • $\begingroup$ Thank you so much for clarifying that. :) $\endgroup$ – Silent Aug 8 '15 at 9:34
  • $\begingroup$ I have written a competing answer below. Just pinging since I felt you might be interested in glancing through it. $\endgroup$ – anonymous_user Jan 15 at 22:48
  • $\begingroup$ A little bit more rigorous would be that the area of the triangle with sides $(\sec\theta, \tan(\theta + d\theta) - \tan \theta , \sec(\theta+d\theta)) = \frac 12 (\tan(\theta + d\theta) - \tan \theta)$ And, this is greater than the area of the section of the circle of radius $\sec\theta$ and less than the section of the cricle with radius $\sec(\theta + d\theta)$ Establishing the inequality $\sec^2\theta \ d\theta \le \tan(\theta + d\theta) \le \sec^2 (\theta + d\theta)\ d\theta.$ We can the use the squeze theorem to find $\frac {d}{d\theta}\tan\theta$ $\endgroup$ – Doug M Jan 15 at 22:56
  • $\begingroup$ @anonymous_user: Thanks for the ping, see my comment on your answer for why it's wrong. $\endgroup$ – user21820 Jan 16 at 7:41
  • $\begingroup$ @DougM: As you and I both know, there are many ways to rigorously obtain the desired theorem. But handwaving in the form quoted in the question is not one of them. By the way, if you want proper mathematical rigour, you should not invoke area unless you have rigorously defined area and proven the theorems required to justify your "a little bit more rigorous" argument. I guess you probably knew that too. $\endgroup$ – user21820 Jan 16 at 7:44

By "ultimately similar" Needham means the following (see pages 20–21):

We stress here the words "ultimately equal" and "infinitesimal" are being used in delicate, technical senses; in particular, "infinitesimal" does not refer to some mystical, infinitely small quantity7. More precisely, if two quantities $X$ and $Y$ depend on a third quantity $\delta$, then \begin{align*} \lim_{\delta \to 0} \frac{X}{Y}\quad & \iff \quad \text{"$X = Y$ for infinitesimal $\delta$".}\\ &\iff \quad \text{"$X$ and $Y$ are ultimately equal as $\delta$ tends to zero".} \end{align*} It follows from the basic theorems on limits that "ultimate equality" inherits many of the properties of ordinary equality.

7For more on this distinction, see the discussion in Chandrasekhar [1995].

Since two triangles are similar if and only if they have the same set of (internal) angles when Needham says that "the black triangle is ultimately similar [exercise] to the shaded triangle" he probably means that the ratio of any angle of the black triangle with the corresponding angle of the shaded triangle tends to $1$ as $d\theta$ tends to zero.

Needham's diagram but with labels added to the vertices.

Let the shaded triangle be labelled $\triangle \mathit{ABC}$, where $\angle A = \theta$ and $\angle B = \pi/2$. The black triangle $\triangle\mathit{CDE}$ is obtained by constructing the perpendicular to the line segment $\mathit{AC}$ at $C$ and taking the point $D$ to be where it meets the straight line that is at an angle of $\theta + d\theta$ to the base $\mathit{AB}$, and by extending the segment $\mathit{BC}$ until it meets the same line at the point $E$.

Let us compute the angles of $\triangle \mathit{CDE}$. Examining $\triangle \mathit{ABE}$, we have $\angle E = \pi/2 - \theta - d\theta$ in $\triangle \mathit{CDE}$. Examining $\triangle \mathit{ACD}$, we have $\angle D$ in this triangle to be $\pi/2 - d\theta$. Hence, $\angle D$ in $\triangle \mathit{CDE}$ equals $\pi/2 + d\theta$. Thus, we must have $\angle C$ in $\triangle \mathit{CDE}$ to be equal to $\theta$ (this could also have been seen directly from our construction of $\triangle \mathit{CDE}$).

So, what do the ratios of corresponding angles between $\triangle \mathit{ABC}$ and $\triangle{\mathit{CDE}}$ look like in the limit as $d\theta \to 0$? \begin{align*} &\lim_{d\theta \to 0} \frac{\angle \mathit{CAB}}{\angle \mathit{DCE}} = \lim_{d\theta \to 0} \frac{\theta}{\theta} = 1.\\ &\lim_{d\theta \to 0} \frac{\angle \mathit{BCA}}{\angle \mathit{CED}} = \lim_{d\theta \to 0} \frac{\pi/2 - \theta}{\pi/2 - \theta - d\theta} = 1.\\ &\lim_{d\theta \to 0} \frac{\angle \mathit{ABC}}{\angle \mathit{EDC}} = \lim_{d\theta \to 0} \frac{\pi/2}{\pi/2 + d\theta} = 1.\\ \end{align*}

This proves that the black triangle and shaded triangle in Needham's diagram are ultimately similar. (Of course, we could have just done this for any two angles, since the third angle of a triangle is uniquely determined once the other two are specified.)

Zoom in on the black triangle.

Now, call $\lvert \mathit{CE} \rvert$ to be $dT$. What can we say about $\lvert \mathit{CD} \rvert$? Let $F$ be the point on the segment $\mathit{AD}$ such that $\lvert \mathit{AF} \rvert = \lvert \mathit{AC} \rvert = L$. Then, certainly $\lvert \mathit{CF} \rvert < L\, d\theta < \lvert \mathit{CD} \rvert$. So, to show that $L\, d\theta$ and $\lvert \mathit{CD} \rvert$ are ultimately equal, it suffices to show that $\lvert \mathit{CF} \rvert$ and $\lvert \mathit{CD} \rvert$ are ultimately equal.

Applying the cosine law on $\triangle \mathit{ACF}$, we get $$ \cos(d\theta) = \frac{L^2 + L^2 - \lvert \mathit{CF} \rvert^2}{2^{\vphantom{2}}\!\;\! L^2} \implies \lvert \mathit{CF} \rvert = 2L\sin\left(\frac{d\theta}{2}\right). $$ Examining $\triangle \mathit{ACD}$, we have $$\tan(d\theta) = \frac{\lvert \mathit{CD} \rvert }{ L } \implies \lvert \mathit{CD} \rvert = \frac{2L\tan \left(\frac{d\theta}{2}\right)}{1^\vphantom{2}\! - \tan^2 \left(\frac{d\theta}{2}\right)}. $$ Thus, $$ \frac{\lvert \mathit{CF} \rvert}{\lvert \mathit{CD} \rvert} = \cos\left(\frac{d\theta}{2}\right) \cdot \left(1 - \tan^2\left(\frac{d\theta}{2}\right)\right). $$ As $d\theta$ tends to $0$ the RHS tends to $1$, so $\lvert \mathit{CF} \rvert$ and $\lvert \mathit{CD} \rvert$ are ultimately equal.

To summarise, the ratio $\dfrac{\lvert \mathit{CE} \rvert}{\lvert \mathit{CD} \rvert}$ is ultimately equal to $\dfrac{\lvert \mathit{AC} \rvert}{\lvert \mathit{AB} \rvert}$ because $\triangle \mathit{CDE}$ is ultimately similar to $\triangle \mathit{ABC}$, and $\lvert \mathit{CD} \rvert$ is ultimately equal to $L\, d\theta$. Thus, $\dfrac{dT}{L\, d\theta}$ is ultimately equal to $\dfrac{L}{1}$, so we get $$ \frac{dT}{d\theta} = L^2 = 1 + T^2 $$ in the limit as $d\theta \to 0$.

The reference to Chandrasekhar [1995] mentioned in footnote 7 in the quote at the beginning is listed on page 574 as follows: S. Chandrasekhar (1995), Newton's Principia for the Common Reader. Clarendon Press, Oxford.

  • $\begingroup$ I think you meant to have "$\cdots{}=1$" in your first limit expression. However, your argument is wrong, and illustrates why one should do proper real analysis rather than using imprecise handwaving about "ultimately equal". At one point you effectively claim that if $a<b·d<a+c$ and $c → 0$ as $d → 0$, then $b·d/a → 1$ as $d → 0$. This is false! You did not control the rate at which $a → 0$ as $d → 0$. I did not continue checking past that error. $\endgroup$ – user21820 Jan 16 at 7:40
  • $\begingroup$ @user21820 Ah, quite right... Thank you for going through it and giving feedback. Indeed, there is no point in checking past that error. $\endgroup$ – anonymous_user Jan 16 at 9:53
  • $\begingroup$ @user21820 I have tried to fix the argument now. $\endgroup$ – anonymous_user Jan 16 at 13:59
  • $\begingroup$ I'm sorry I missed an earlier gap in your argument. You claimed that $L·dθ < |CD|$, but this is very difficult to justify rigorously. In fact, the other inequality $|CF| < L·dθ$ is also not justified but at least can reasonably be hand-waved (though a rigorous justification would still have to define arc length of a smooth curve). I hope you realize now that it is highly non-trivial to rigorously prove the desired result along this kind of line of reasoning. In case you are interested, see this post and the post that it links to. $\endgroup$ – user21820 Jan 16 at 15:15
  • $\begingroup$ @user21820 Thank you once again for taking the time to go through my answer and giving feedback. If I'm not mistaken, the claim that $L\, d\theta < \lvert \mathit{CD} \rvert$ is implied by (or equivalent to?) $x < \tan(x)$ for $0 < x < \pi/2$, and $\lvert \mathit{CF} \rvert < L\, d\theta$ is saying that the length of a chord in a circle is less than the length of the arc subtended by that chord. $\endgroup$ – anonymous_user Jan 16 at 21:00

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