(x+yi)^i how?!

[Nelde95]

As the title suggests, I am wondering about how you take a complex number to the power of i.
It would be preferable if the result did not involve infinite series or numbers with an infinite number of decimals, so the result can be written in a finite amount of space on the paper, while still being 100% accurate...
I am asking this here because this is the only forums I am a member of.
Hope that it is obvius that if i is still in the exponent I am just as lost as before.
Tried to google it but I did not find the answer I was looking for.

[Jenkar]

If x=0 & y = 0, your result is 0.

Else, it can be written as (re^i(∂))^i.
That's r^i * e^(-∂)
r = e^ln(r)
So your result is e^(i*ln(r)) * e(-∂)

This is a bet but i think it's pretty solid.

[Nelde95]

If x=0 & y = 0, your result is 0.

Else, it can be written as (re^i(∂))^i.
That's r^i * e^(-∂)
r = e^ln(r)
So your result is e^(i*ln(r)) * e(-∂)

This is a bet but i think it's pretty solid.

Studying imaginary numbers on my own, could you explain "∂".
also when I look at it I get i in the exponent...

Hope that it is obvius that if i is still in the exponent I am just as lost as before.

Edit:
I must also admit that my knowledge stops at about sqrt(x+yi) wich I found out on my own.

[RavingRabbid]

Consider an angle of 1 radiant.

Rotate (x+yi) by 1 radiant.

(or, it should be)



(x cos 1 - y sen 1) + i (x sen 1 + y cos 1)

Or I'm on drugs.

[bogtro]

Note that e^(ix)=cos(x)+isin(x), more commonly known as cis(x).

As a consequence, cis^n(x)=cis(nx)

For example,

i^i=(cis pi/2)^i=cis^i(pi/2)=cis(i*pi/2)

Since cis(x)=e^(ix),
cis(i*pi/2)=e^(-pi/2).

This is as accurate a result as is possible.

[Jenkar]

There's something that says that any complex that is not 0 can be put on the form re^(i∂), ∂ being a real and r too. There's a theorem (proven, but i have never seen the proof nor probably have the level to understand it) that says that e^(i∂) = cos(∂) + i * sin(∂).
hence the e^(i * ln(r)) in the solution is cos (ln(r)) + i * sin(ln(r)).

Square roots are fun. want me to explain that too?

[Nelde95]

Woah, all of this takes a lot of braincapacity to swallow. think ill read it all tomorow.
I said that I figured out the squareroots my self:
ReOut^2-ImOut^2=ReIn
2*ReOut*ImOut=ImIn

Edit:
Failed because I am tired...
Forgot to putt ImOut to the power of 2 in the first equation, thanks for pointing it Out. <- ucwhatididthere?

[bogtro]

The theorem you are referring to is Euler's Theorem (or Formula). The proof is relatively complex and requires a knowledge of calculus, specifically Taylor series. The first statement follows from this theorem.

[Jenkar]

Actually,
ReOut^2 - ImOut^2 = ReIn
2*ReOut*ImOut = ImIn

And it's very useful to see that :
ReOut^2 + ImOut^2 = sqrt(ReIn^2+ImIn^2)

[bjessee]

All complex numbers x+yi can be written in polar notation re^i∂ or r(cos ∂ + isin ∂).

r in that sense is the magnitude of (x+yi) or sqrt(x^2 + y^2).  The weird symbol, ∂, denotes the angle atan2 (y/x).

Using the identity x = e^ln x, we can simplify (x+yi)^i as e^ln [(x+yi)^i] or e^(i*ln (x+yi)).  Now sub in re^i∂ for (x+yi) giving e^(i*ln re^i∂) or e^i(ln r + ln e^i∂) or e^(iln r - ∂).  This can be further simplified to (e^-∂)(e^i*ln r) or (e^-∂)[cos (ln r) + isin (ln r)].

You can substitute back for r and ∂ if you wish, but it is probably cleaner to leave it as is.

Hope this helps.

[RavingRabbid]

m = sqrt(x^2+y^2)

t = arctan(y/x)


(x+iy)^i = (m(e^(it)))^i = m^i * e^-t = e^i(ln m) * e^-t

[Nelde95]

I am comfused with what everyone means with i, is it refering to the imaginary part of a number or just i? I llike using the notation ImIn, ImOut, ReIn and ReOut, to clarify.

Also: might be good to note that i am born in 95 and not 59. (Did not try the game "Riviera" before yesterday, so my user name is not refering to the ruins of nelde. It is an old username backwards. If it was posible to rename both forum name and ign it would be nelde95).