Euler's Formula

By Neil Morrison on Tuesday, November 23, 1999 - 11:03 am : I already know the series proof of the more general de Moivre formula; I merely used the version with

π

as an example. This formula has led me to some interesting conclusions!

$e^{x}$ is periodical (period $2 π i$ )
hence $e^{y} = x$ has an infinite number of values of $y$ for each $x$ .
$\ln x = \ln | x | + i \times \arg (x)$
eg: $\ln i = 0 + i \times π / 2$
$\ln (- 1) = 0 \pm i \times π$
$i^{i}$ is a real number!
$i^{i} = (e^{\ln i})^{i} = e^{i \ln i} = e^{i i (π / 2)} = e^{- (π / 2)} = 0.208 \dots$

I would be grateful if you would elaborate on these conditions, and point out any flaws, and also advise me what other areas I can analyse using complex numbers.

Thanks

Yours,

Neil Morrison

By Matt Daws (Mdpd2) on Tuesday, November 23, 1999 - 11:32 am :

You point out, correctly, that

e^{x + 2 π n i} = e^{x}

for any integer

n

. Hence, if we define

\ln x

to be the inverse of

e^{x}

, we have that

\ln x = \ln | x | + i \times \arg (x) + 2 π n i

for any

n

. This is just another way of pointing out that the argument function,

\arg (x)

, is not well defined: we can add

2 n π

to it and it will still behave the same.

Usually you just define $\arg (x)$ to be in a certain range, say $0 \leq \arg (z) < 2 π$ , and have done with it. However, this can lead to problems: consider a loop going about the origin, and what the value of $\arg (z)$ will be on that loop. Using my definition, it'll jump between 0 and $2 π$ at some point, even if the loop is very well behaved (e.g. look at the circle about the origin, moving in an anti-clockwise direction: $\arg (z)$ will increase from 0 to $2 π$ but will then jump back to 0).

This sort of behaviour also means that any definition of $\ln z$ will jump like this, which is not something we want. You can get around it by only defining $\ln z$ on certain sections of the complex plane: i.e. only for certain values of $z$ , so that we can't get this jumping problem.

I don't know if you are at school or university or whatever: here at Cambridge the courses Complex Analysis and Further Analysis deal with these issues: a much more complex way of dealing with it is to use Riemann surfaces, which in some manner allow you to define $\arg (x)$ to be all possible values.

By Neil Morrison (P1462) on Tuesday, November 23, 1999 - 06:20 pm :

Thanks for your prompt reply! Today, I also worked out (3+4i)²⁺²ⁱ as a test. As I am doing CSYS Maths (Papers 1&2), this was just some personal research. I found an internet page (forgotten where) that gave Euler's formula, and some properties to investigate.

By Matt Daws (Mdpd2) on Wednesday, November 24, 1999 - 12:18 am :

Yeah, it's an interesting area: Complex numbers behave very differently to Real numbers in some respects. For example, it is approximately true to say that any integral in the complex numbers doesn't depend on the path taken.

Path? In the real line, there is only one way to get between any two numbers. In the complex plane, there are loads of ways, so to integrate, you have to say what line you're integrating along for it to make any sense. Luckily, it turns out that it actually doesn't really matter! That's quite an odd idea, if you think about it for a bit.

Anyway, I'm not sure I can do a good job of saying much more without confusing you utterly. Maybe someone else has some ideas? There are some good books around, but most probably assume more than you know (e.g. about formal real analysis/calculus), so probably won't be very useful.

By Andrew Wyld (Acew2) on Friday, November 26, 1999 - 01:36 am :

I'd like to add a comment to the above.

Normally it doesn't matter what path you choose for integration, no, but suppose you are taking the definite integral of some function between two points. Let's say its indefinite integral is $\arg (z)$ plus some constant (this works ok, and saves me having to define the actual function we're integrating, which is tedious and irrelevant). The definite integral is well-defined, we hope, since the constant drops out. So what you end up with in your integral is the difference between the arguments of two points.

HOWEVER.

Let's choose the points 1 and -1 , for ease. If you go above the origin with your path of integration, $\arg (z)$ increases along the path and so $\arg (- 1)$ is $π$ plus some constant, $\arg (1)$ is zero plus the same constant, and hence the difference between the two points - and hence the value of the integral - is $π$ . Take a path below the origin, however, and the value of $\arg (z)$ decreases. Then $\arg (- 1)$ is $- π$ plus some constant, $\arg (1)$ is still zero plus the same constant, but your integral now has the value $- π$ . Weird.

This is because the origin (which is the centre of all this trouble) is a singularity for $\arg (z)$ - which is to say that $\arg (z)$ is not well-defined there (worse things can happen, but that's all you need, really)! This is obvious when you think about it. Whatever angle you choose, the point will be the same.

So although you can choose any path of integration for complex numbers, if you pass through a singularity (where the function is ill-defined) expect the value of your integral to change.

Interesting thought - I haven't covered all the possibilities for $\arg (z)$ - what about multi-spiral paths? This is where Riemann surfaces come in - you think of yourself as being on a "different incarnation" of the complex plane, so to speak, every time you sprial round, like a corkscrew shape. There's a lot more to it than that. None of which I know.

I'll shut up now.

Laters,

Andrew Wyld

By Theo Zapata on Saturday, November 27, 1999 - 11:59 pm :

If you wanna see a good problem using the Moivre's formula.... calculate this sum.

[\cos (π / 4)] / 2 + [\cos (2 π / 4)] / 2^{2} + \dots + [\cos (n π / 4)] / 2^{n}