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I'm reading the answer here (https://math.stackexchange.com/a/906108/1013993), and the following derivation is made:

We start with

$$\Delta z = r\,\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)} - r \,\mathrm{e}^{\mathrm{i}\theta}$$

and then use the derivative

$$\mathrm{e}^{\mathrm{i}(\theta+\Delta\theta)}-\mathrm{e}^{\mathrm{i}\theta} = \frac{\mathrm{d}\mathrm{e}^{\mathrm{i}\theta}}{\mathrm{d}\theta}\Delta\theta = \mathrm{i}\mathrm{e}^{\mathrm{i}\theta}\Delta\theta \quad(1)$$

to conclude

$$\Delta z = \mathrm{i}r\,\mathrm{e}^{\mathrm{i}\theta}\Delta\theta.$$

I presume that here they are using the limit definition, that is,

$$\frac{de^{i\theta}}{d\theta}\Delta \theta = \lim_{\Delta \theta \to 0}\frac{e^{i(\theta + \Delta \theta)} - e^{i \theta}}{\Delta \theta}\Delta \theta$$

in which case, I'm wondering where the $\lim$ term goes, since they seem to drop it in equating the LHS and the middle term in (1).

1 Answers1

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As geetha290krm indicates, as written, these are not exact equations. Possibly syockit didn't know how to do a $\approx$ sign in LaTeX. Or maybe they were just being a little lazy and expected readers to recognize it as an approximation instead of a strict equality.

What they meant was

For small $\Delta\theta$, $$e^{i(\theta+\Delta\theta)}-e^{i\theta} \approx \frac{de^{i\theta}}{d\theta}\Delta\theta = ie^{i\theta}\Delta\theta $$

This is a sort of "rough-and-ready" calculation technique that dates back to before Cauchy and Weierstrass inserted true mathematical rigor into calculus. It is still common in applied mathematics, particularly among physicists, who are given to simply assuming their functions are regular enough to not cause issues. Note that the reference is to a book for "Engineers and Scientists".

The limit as $\Delta\theta \to 0$ is introduced later, when they derive the formula for $f'(z)$. It is to be understood that the approximation is sufficient for the limit calculation to be true. Which it is. This basically an application of the chain rule.

Paul Sinclair
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  • Though later on, others added rigor for the infinitesimal approach in nonstandard analysis. – Alan Sep 08 '22 at 20:54
  • @Alan - but not in the calculation under discussion here, where $\Delta \theta$ was not treated as an infinitesimal. – Paul Sinclair Sep 08 '22 at 22:37