# What is wrong?

• January 7th 2006, 01:14 PM
ThePerfectHacker
What is wrong?
I posed to myself two problems:
$1^{\pi}=?$
and
$(-1)^{\pi}=?$.
For the second one I did $(-1)^{\pi}=(e^{\pi i})^{\pi}=\cos (\pi^2)+ i\sin(\pi^2)$ it seems good.
But for the second one
$1^{\pi}=(e^{2\pi i})^{\pi}=\cos (2\pi^2)+i\sin (2\pi^2)$ yet it should be one!?!?

I am guessing the exponent for complex numbers is defined for $e^{\theta i}$ where $0\leq \theta<2\pi$ Thus, my manipulation in problem 2 is out of the domain of the definition and it causes problems.
Am I right?
• January 7th 2006, 01:32 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I posed to myself two problems:
$1^{\pi}=?$
and
$(-1)^{\pi}=?$.
For the second one I did $(-1)^{\pi}=(e^{\pi i})^{\pi}=\cos (\pi^2)+ i\sin(\pi^2)$ it seems good.
But for the second one
$1^{\pi}=(e^{2\pi i})^{\pi}=\cos (2\pi^2)+i\sin (2\pi^2)$ yet it should be one!?!?

I am guessing the exponent for complex numbers is defined for $e^{\theta i}$ where $0\leq \theta<2\pi$ Thus, my manipulation in problem 2 is out of the domain of the definition and it causes problems.
Am I right?

$1=e^{2.n.\pi .i}$, and $-1=e^{(2.n+1).\pi .i}$, for $n=0, \pm 1, \pm 2, ..$

Now raise these to the power $\pi$, and you have multiple values one
of which will probably be what you are expecting.

RonL
• January 7th 2006, 05:07 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
$1=e^{2.n.\pi .i}$, and $-1=e^{(2.n+1).\pi .i}$, for $n=0, \pm 1, \pm 2, ..$

Now raise these to the power $\pi$, and you have multiple values one
of which will probably be what you are expecting.

RonL

Thus, you are saying there are infinitely many answers?
• January 8th 2006, 02:26 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Thus, you are saying there are infinitely many answers?

Yes, if we alow complex values.

RonL
• January 9th 2006, 03:01 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
Yes, if we alow complex values.

RonL

Then explain this to me:
$1^{\pi}=(e^{2\pi i})^{\pi}=\cos{2\pi^2}+i\sin{2\pi^2}$
And,
$1^{\pi}=(e^0)^{\pi}=\cos0+i\sin0$
Thus,
$1=\cos0+i\sin0=\cos{2\pi^2}+i\sin{2\pi^2}$
But that is not true ?!?!
• January 10th 2006, 04:21 AM
TD!
You have to be careful with complex numbers, not all the properties you know from the real numbes (about square roots, powers, logarithms, ...) still hold for complex numbers.

In particular, the rule $(z^a)^b = z^{ab}$ no longer holds.
The correct version is $(z^a)^b = z^{ab}e^{2bk\pi i}$.

So $(e^{2\pi i})^{\pi} = e^{2\pi^2 i}e^{2k\pi^2 i} = e^{2\pi^2 i+2k\pi^2 i} = e^{(k+1)2\pi^2 i}$

And this again yields $e^0$ for $k = -1$.

This has to do with the analytical continuation of the natural logarithm as a complex function and the fact that for complex values of z, $e^z$ is a periodic function with period $2\pi i$.
• January 10th 2006, 12:03 PM
ThePerfectHacker
Quote:

Originally Posted by TD!
You have to be careful with complex numbers, not all the properties you know from the real numbes (about square roots, powers, logarithms, ...) still hold for complex numbers.

In particular, the rule $(z^a)^b = z^{ab}$ no longer holds.
The correct version is $(z^a)^b = z^{ab}e^{2bk\pi i}$.

So $(e^{2\pi i})^{\pi} = e^{2\pi^2 i}e^{2k\pi^2 i} = e^{2\pi^2 i+2k\pi^2 i} = e^{(k+1)2\pi^2 i}$

And this again yields $e^0$ for $k = -1$.

This has to do with the analytical continuation of the natural logarithm as a complex function and the fact that for complex values of z, $e^z$ is a periodic function with period $2\pi i$.

I knew that all of these "paradoxes" I was getting was the result of incorrect rules for complex numbers. It is like the square root of the product of two negative numbers IS NOT the product of the square roots. I just never studied complex analysis.
• January 10th 2006, 12:05 PM
TD!
Quote:

Originally Posted by ThePerfectHacker
I knew that all of these "paradoxes" I was getting was the result of incorrect rules for complex numbers. It is like the square root of the product of two negative numbers IS NOT the product of the square roots. I just never studied complex analysis.

Indeed, that's also one of the consequences. It follows from the fact that when you analytically continue the square root function as a complex function, you'll have to introduce a branch cut and this makes it possible for the 'old rule' we had for the real case to give false results in the complex case.
• January 10th 2006, 09:29 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Then explain this to me:
$1^{\pi}=(e^{2\pi i})^{\pi}=\cos{2\pi^2}+i\sin{2\pi^2}$
And,
$1^{\pi}=(e^0)^{\pi}=\cos0+i\sin0$
Thus,
$1=\cos0+i\sin0=\cos{2\pi^2}+i\sin{2\pi^2}$
But that is not true ?!?!

If you replace the $\pi$ exponent by $1/2$ this
line of argument will give $1=-1$ :eek:

RonL