Square root of minus one

If i^2= (-1), what is i^i? i to the ith power.

i^i = exp(-PI/2+2.n.PI)

What is n in that equation?
And how does PI enter into this? interesting.
Also, what does exp mean? exponent? if so, what exponent?

Originally posted by sonhouse
What is n in that equation?
And how does PI enter into this? interesting.
Also, what does exp mean? exponent? if so, what exponent?

Originally posted by TheMaster37
He used another way of writing i:

i = e ^ (PI * i / 2)

Since 1 = e ^ (2*PI*i) you can add extra factors of e ^ (2*PI*i), hence the n.

So i^i = e ^(-PI/2) mod e ^ (2*PI)

Originally posted by AThousandYoung
i is part of the definition of i? That doesn't make sense.

Originally posted by mtthw
It's not part of the defiition of i, but it's an extremely useful general result:

cos(theta) + i sin(theta) = e^(i theta)

He's using the case where theta = PI/2. Then the left-hand side is equal to i.q

You just use the fact that:

e^(b.i)= cos(b)+i. sin(b) (1)

This is equal to i (look at the RHS) when b = PI/2 - (or +) 2.n.PI with n being any integer.

i^i = e^[(b.i)^i]= e^(-b) = e^ (PI/2 - 2.n.PI).i.i = e ^ (-PI/2 + 2.n.PI)

Originally posted by mtthw
It's not part of the defiition of i, but it's an extremely useful general result

Originally posted by Palynka
Yep, someone that needs to deal with dynamic optimization should be familiar with it. It's called de Moivre's theorem.

Originally posted by mtthw
Or almost any serious applied mathematics. It crops up in wave theory, quantum mechanics, fluid dynamics, instability calculations...

Calculating with i ( as sqrt(-1) ) is only an extension of the real number system. What you can do with R you can do with C.

...and more:

With the complex number system (C) you can calculate square root of any number (including complex numbers), not only positive numbers. You can calulate arcsin with any number (including complex numbers), not only numbers between -1 and 1. You can take logaritm of any number (including complex numbers), not only positive numbers. And so on.

With i = sqrt(-1) you can calculate i^i, sqrt (i), log (i) with 'no problem'. But you cannot calculate i/0 but who can?

But one pecularity of the complex number is that no complex number is larger or smaller than any other number. Another one is that negative infinity is the same as positive infinity.

Complex numbers are fun!

Originally posted by sonhouse
Nice. Thanks for that. I Goog'd De Moivre and found this neat wiki:
http://en.wikipedia.org/wiki/De_Moivre's_formula
This piece goes into it in some detail.
BTW, my present read is 'The Riemann Hypothesis"
by Karl Sabbagh. Anyone read this? Great read if you are into maths.
I think my next maths book will be 'the joy of i'. I think there is a book with something like that title.

But one peculiarity of the complex number is that no complex number is larger or smaller than any other number. Another one is that negative infinity is the same as positive infinity.

I follow everything but that one. It sounds like you are saying all complex numbers are equal?? And that (+inf) + (-inf)=0? and (+inf) * (-inf)= (-inf)^2?

Originally posted by FabianFnas
Calculating with i ( as sqrt(-1) ) is only an extension of the real number system. What you can do with R you can do with C.

...and more:

With the complex number system (C) you can calculate square root of any number (including complex numbers), not only positive numbers. You can calulate arcsin with any number (including complex numbers), not only ...[text shortened]... er one is that negative infinity is the same as positive infinity.

Complex numbers are fun!