e^ipi squared

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January 8th 2009, 08:40 AM
hmmmm

e^ipi squared

today in class i was asked what was wrong with this
e^ipi + 1=0
e^ipi=-1
e^2ipi=1
2ipi=ln1
this is obviously wrong, is it the stage where you square e^ipi,
i havnt come across complex powers before. thanks for any help sorry if this is a stupid question.
January 8th 2009, 10:19 AM
shinhidora

Quote:

Originally Posted by hmmmm

today in class i was asked what was wrong with this
e^ipi + 1=0
e^ipi=-1
e^2ipi=1
2ipi=ln1
this is obviously wrong, is it the stage where you square e^ipi,
i havnt come across complex powers before. thanks for any help sorry if this is a stupid question.

you mean?

$e^{\pi} + 1 = 0$

if you mean this, then it's obviously wrong because it isn't 0 xD

$e^{\pi} + 1 = 24.14069263...$
January 8th 2009, 10:22 AM
masters

Quote:

Originally Posted by hmmmm

today in class i was asked what was wrong with this
e^ipi + 1=0
e^ipi=-1
e^2ipi=1
2ipi=ln1
this is obviously wrong, is it the stage where you square e^ipi,
i havnt come across complex powers before. thanks for any help sorry if this is a stupid question.

Hello hmmmm,

See here: Euler's identity - Wikipedia, the free encyclopedia
January 8th 2009, 10:24 AM
Moo

No, $e^{i \pi}+1=0$ and that's true :)

The problem is when you take the logarithm.
Because $e^{i \alpha}=e^{i (\alpha+2k \pi)}$ where k is an integer, this is why your step is wrong.

So following this reasoning, you can have $e^{2i \pi}=e^{i(2 \pi-2\pi)}=e^0=1$ which is indeed true.

You can have a look at this : Complex logarithm - Wikipedia, the free encyclopedia
January 8th 2009, 10:56 AM
hmmmm

thanks

thanks but if possible could you elaborate on your third line?
January 8th 2009, 11:05 AM
Moo

Quote:

Originally Posted by hmmmm

thanks but if possible could you elaborate on your third line?

Imagine that $e^{i \alpha}=\cos(\alpha)+i \sin(\alpha)$ (this is a definition)
Since the cosine and the sine are $2 \pi$ -periodic functions, we can write :
$e^{i \alpha}=\cos(\alpha)+i \sin(\alpha)=\cos (\alpha+2k \pi)+i \sin(\alpha+2k \pi)=e^{i (\alpha+2k \pi)}$
The logarithm in the complex values has an infinity of values, unless you define a continuous logarithm (but I don't think this is something you need to know yet).

another way of seeing it is that $1=1+i 0=\cos(2 k \pi)+i \sin (2 k \pi)=e^{i 2k \pi}$
So you can multiply $e^{i \alpha}$ by $e^{i 2 k \pi}$ , which gives $e^{i (\alpha+2k \pi)}$ , it won't change the value.
January 8th 2009, 11:14 AM
Soroban

Hello, hmmmm!

Quote:

Today in class i was asked what was wrong with this:

. . $e^{i\pi} + 1\:=\:0$

. . $e^{i\pi} \:=\:-1$

. . $e^{2i\pi}\:=\:1$ . . . square both sides

. . $2i\pi\:=\:\ln1 \:=\:0$

This is obviously wrong.
Is it the stage where you square $e^{i\pi}$ ? . . . . yes

We have: . $e^{i\pi} \:=\:-1$

Square: . $e^{2i\pi} \:=\:(-1)^2$

Then: . $2i\pi \:-=\:\ln(-1)^2 \:=\:2\ln(-1)$

Hence: . $i\pi \:=\:\ln(-1)$ . . . and this is true.
January 8th 2009, 11:17 AM
Moo

Quote:

Hence: . $i\pi \:=\:\ln(-1)$ . . . and this is true.

I'm sorry to say, but it's not always true. It depends on the logarithm you take.

You can define a logarithm such that $\ln(-1)=3i \pi$
Read my explanation (or what it tries to be) above.