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Value of \((ln(1))^2 \)

Where did i go wrong ? \((ln1)^2 = (ln(e^{i2\pi}) )^2 = -4(\pi)^2 \)

Note by Kushagraa Aggarwal
4 years, 2 months ago

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From one point of view, you didn't. It is impossible to define the logarithm continuously on the whole complex plane. Basically we have to have \[ \ln z \; = \; \ln|z| + i\mathrm{Arg}(z) \qquad \qquad z \neq 0\] and the problem lies in the fact that \(\mathrm{Arg}(z)\) is not continuously well-defined on the whole complex plane, or even on the complex plane with \(0\) removed.

A choice of convention is called for. The standard definition, or "principal branch", of the logarithm would restrict its definition to the cut plane of the complex numbers with the negative real axis removed, so that \(|z|>0\) and \(-\pi < \mathrm{Arg}(z) < \pi\) throughout. Then the above definition works fine, and has the advantage that the logarithm of positive real numbers coincides with the normal definition for positive reals. This definition of the logarithm is analytic on its cut plane.

However, you could choose a branch of the logarithm which was defined on the complex plane with the positive imaginary axis removed, with the assumed convention for Argument being such that \[ \tfrac{\pi}{2} < \mathrm{Arg}(z) < \tfrac{5}{2}\pi \] throughout the region. In this case, the value of \(\ln 1\) would indeed be \(2\pi i\). This definition of logarithm would also be analytic in its cut plane.

It is simply a matter of convention and a wish for consistency that makes us choose the first definition of the logarithm whenever possible.

Mark Hennings - 4 years, 2 months ago

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It's possible to go a bit better than this -- in fact, this gets to the heart of why complex numbers are so awesome.

Instead of restricting ourselves to the complex plane, and introducing an artificial discontinuity, we can imagine the domain of \(\log z\) to potentially be some surface other than a plane. But it has to have the property that the function varies smoothly over the extent of the surface.

(As an aside, we can also do this with \(f(z) = 1/z\). That function naturally tends to have a sphere for the domain - with the south pole being 0 and the north pole being \(\pm \inf\)). The function maps the sphere onto an upsidedown copy of the sphere.

If we consider taking a little circle around the point \(z = 0\), and evaluating the \(\log\) function at each point, we will find that once we complete the circle we don't get back to where we started.

Specifically, consider \(\log{e^{(0.1)i}}\), \(\log{e^{(0.2)i}}\), ...., until we get up to \(\log{e^{(2\pi)i}} = 2\pi i\). We went around \(0\) in a circle but we now find ourselves in a different place.

In other words, taking a curved path that turns through \(2\pi\) radians, if we integrate along that path we actually don't get 0 as we would for a real function of two variables; we get \(2 \pi i\).

So we can think of the "full" domain of \(\log z\) to be like a spiral staircase, with the central pillar at \(z = 0\). If we take this as the domain then the function is smooth everywhere, except for a finite number of discontinuities - in fact, just \(z=0\) in this case.

Here's a visualization: http://en.wikipedia.org/wiki/File:Riemannsurfacelog.jpg

Matt McNabb - 4 years, 1 month ago

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Nice post Mark H.! Thank you.

Pranav Arora - 4 years, 2 months ago

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\(\Im(\text{Santa Claus})=\text{Santa Claus}\)

Cody Johnson - 4 years, 2 months ago

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Ho ho ho, Of course, since \(x+iy=y \Rightarrow x=y=0\) for real \(x,y\), if Santa Claus were equal to his imaginary part, he would be \(0\), so real, and not imaginary.

Mark Hennings - 4 years, 2 months ago

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its like saying e^{0}=e^{2i(pi)} means 2i(pi)=0 or i=0

Karan Jain - 4 years, 2 months ago

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