Question

Math is so scary.\[\text{irrational}^{\text{unreal} \times {\text{irrational}}} = \text{integer} \]That is really scary(Euler's Identity).
How can we even have something like this?

Answer 1

How can we raise an imaginary power?

Answer 2

To give context, \(e^{i\pi} = -1\)

Answer 3

use Euler's identity ...

Answer 4

Do you search the derivation of Euler's Identity?

Answer 5

Of course, that is Euler's Identity.
Exactly, @Spacelimbus
Just the intuition.

Answer 6

Well, I doubt that there is any intuition to be honest, it's a lot of playing around with series and expansion.

Answer 7

Anything raised to \(i\) is pretty unreasonable, isn't it?

Answer 8

If it would be intuitive, we wouldn't be so fascinated about it right?

Answer 9

Exactly.

Answer 10

well ... i admit this i find myself unintuitive.
http://www.wolframalpha.com/input/?i=i^i

Answer 11

It's surprising how we get no transcendental number. We could still have expected an irrational number, but integer is ridiculous!

Answer 12

what does "unreal" mean?

Answer 13

Imaginary

Answer 14

ooooooh ok

Answer 15

like \(e^{\pi i}\) for example

Answer 16

@satellite73 Lol that is my second reply.

Answer 17

Must things about \(i\) seem very unnatural to me at first, I was fascinated by that already:

\[ \frac{1}{i} =-i \]

Answer 18

That is exactly what I asked. How does this even happen?

Answer 19

@Spacelimbus http://math.stackexchange.com/questions/167502/how-can-i-disprove-the-following-statements

Answer 20

what you have to wrap your head around is what it means to use exponential notation

Answer 21

you know that \(3^2\) means to multiply 3 by itself, but that is not the meaning of \(3^i\) or even \(3^{\sqrt{2}}\)

Answer 22

Yes.

Answer 23

nice link @experimentX

Answer 24

once you define
\[b^x\] as
\[b^x=e^{x\log(b)}\] then you can make more sense out of exponentials, assuming of course that you are familiar with the log

Answer 25

Of course I do, continue.

Answer 26

i was done actually

Answer 27

Oh. Let me see.

Answer 28

but as an example, how would you make sense out of a number like \(2^i\)?

Answer 29

\[2 =b, i = x \Longrightarrow e^{i \ln(2)} \]

Answer 30

yes

Answer 31

But I still find it very surprising to get an integer :P

Answer 32

which is a perfectly good complex number
if you want to write it in standard form \(a+bi\) you can write is as \(\cos(\ln(2))+i\sin(\ln(2))\)

Answer 33

now how about \(e^{\pi i}\) ?

Answer 34

Right.

Answer 35

Wait a sec, thanks.

Answer 36

actually i have to run, but i'll be back later.

Answer 37

\[ a^b = e^{b \ln(a)} \Longrightarrow e^{i \pi} = e^{i\pi \ln(e)} \]

Answer 38

\[\ln(e) = 1 \]

Answer 39

Just simplifies to\[ e^{i \pi}\]:/

Answer 40

\[ \ln e=1 \] But I guess what @satellite73 suggest was using the formula next.

\[ \large e^{i \phi }= \text{cis}(\phi)=\cos \phi + i \sin \phi \]

Answer 41

That's a whole new thing that I learnt today. What is \(\text{cis}\)? Wait. Must post that as a new question.

Answer 42

I don't need the derivation, after all.

Answer 43

cis is just a lazy notation for cos(x) +isin(x)

Answer 44

reads like cosines imaginary sinus.

Answer 45

I'd do that for sure! (By the way, I am thinking about another problem).

Answer 46

often \[ \Large re^{i \varphi} \]
is just defined as :

\[ \Large re^{i \varphi}=r(\cos \varphi +i \sin \varphi)= r\text{cis}\varphi \]

Answer 47

oh ok.