Exercise [05.10] b

vasco · 15 Jun 2008, 07:29

We know that generally $exp(z)=exp(zLog e)=exp(z(\log e+2k\pi i))$ (see my post on the general power w^z

)

It’s important to make the distinction between

1) $exp(z)=exp(zLoge)=exp(z(\log e+2k\pi i))$ which is multi-valued and

2) $exp(z)=1+\frac{z}{1!}+\frac{z^2}{2!}+$ .... which is single valued

Looking at 1) above for z=1 we obtain

$exp(\log e+2k\pi i)=exp(1+2k\pi i)$

Just as (in one of my earlier posts) I used Log w to represent the multi-valued logarithm of w, and log w to represent the principal value, such that
$Log w=\log |w|+i\arg w$ ,
so I could use E(z) to represent the multi-valued exp (z) and e the principal value:

$E(1)=exp(1+2k\pi i)=exp(1)exp(2k\pi i)=e \times exp(2k\pi i)$ , where k is any integer value, positive or negative.

if we take k in the range -2 to +2 then we get:

$E(1)= exp(1-4\pi i)=e\times exp(-4\pi i) \ \ \ \ k= -2$
$E(1)= exp(1-2\pi i) = e\times exp(-2\pi i) \ \ \ \ k= -1$
$E(1)= exp(1)=e \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ k=0$ (the e we all know and love!)
$E(1)= exp(1+2\pi i) = e\times exp(2\pi i) \ \ \ \ k=1$
$E(1)= exp(1 + 4\pi i)=e\times exp(4\pi i) \ \ \ \ k=2$

These are all separate valid values for E(1)

Writing $e=exp(1+2\pi i)$ as is done in Ex 05.10, is not a valid thing to do. It is the same as saying:

$E(1)=exp(1) \ \ \ \ for \ k=0$
$E(1)=exp(1+2\pi i) \ \ \ \ for \ k=1$

And therefore $exp(1)=exp(1+2\pi i)$ .

A simpler, more obvious example of a similar fallacy would be for example sqrt(9). This is also multi-valued and has the 2 possible distinct values -3 and +3.
However, writing

sqrt(9)=+3

and then saying
Therefore +3=-3 is clearly fallacious.

Or, another example

sqrt(9)=+3

but we know sqrt(9)=-3

therefore

fallacious again.

When we say, for example above, that +3=-3 because they are both equal to sqrt(9), we are forgetting that sqrt(9) is a multi-valued function and therefore we are not allowed to equate the two values.

This is exactly the same as saying $E(1) \ for\ k=0$ is equal to $E(1) \ for \ k=1$ .
If we take logarithms then we can see the fallacy immediately:

$\log(E(1)=\log e=1 \ and\ \log(E(1))=1+2\pi i$ and it is clear that
$1 \ne 1+2\pi i$

Exercise [05.10] b

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