Complex indices

1. Sep 28, 2007

JPC

hey, was wondering what would happen if i do :

a^b
with a : real or complex number
and b : a complex number

like : 2^i

2. Sep 28, 2007

genneth

The definition:

$$a^b = \exp (b \ln a)$$

In general ln is multivalued (actually infinitely valued) so there is no unique answer.

3. Sep 28, 2007

JPC

whats exp() ?
i know log() , and ln() but never heard of exp()

4. Sep 28, 2007

edmondng

exponential
e = 2.718281828

5. Sep 28, 2007

genneth

Specifically, $$exp(x) = e^x$$.

6. Sep 29, 2007

JPC

oh ok :

so
2^i = e^ ( i * ln(2))
?

7. Sep 29, 2007

Gib Z

Yup that is perfectly correct. Or if you want it split into real and imaginary parts: $$\cos (\ln 2) + i \sin (\ln 2)$$

8. Oct 24, 2007

JPC

ok
now that we started working on exp(x) in my class i know a bit more what you are talking about

But how do u pass from e^(i * ln(2) ) to $$\cos (\ln 2) + i \sin (\ln 2)$$ ?

9. Oct 24, 2007

SiddharthM

remember that exp(x) and ln(x) are defined in a more general sense as complex functions.
if a complex number is of the form z=a +ib then
exp(z)=exp(a+ ib)=exp(a)exp(ib) by additivity of exponentials and then using euler's formula: exp(z)=exp(a)(cosb +isin(b))
here the a is real so exp(a) is evaluating using the real definition.

The complex definitions of the trancesendental functions extend that of it's real analogue.

JPC: if you learning about the real exponential function for the first time you probably won't encounter euler's formula until later in the course - it's an application of power series.

10. Oct 25, 2007

Gib Z

JPC, it seems my assumptions on your maths education are wrong, in Australia for some reason we do complex numbers after Exponential functions, forgive the pun but here they are seen as, well, more complex. Anyway, You obviously don't need to know what I say from here, so you can either choose to ignore my post completely or read more on what I say next: There is a famous relation that Euler derived through Taylor Series expansions that told him that $$e^{ix} = \cos x + i \sin x$$. I Merely used that identity straight off.

11. Sep 13, 2010

bchui

For $$b=x+iy$$ , $$a^b=a^{x+ i y}=a^x a^{i y} = a^x \exp(i y \ln (a)) = a^x (\cos (y \ln (a))+i \sin (y \ln (a))$$
by Euler's formula $$e^{i\theta}=\cos\theta+i\sin\theta$$.

I wonder what happens to $$A^B$$ where both $$A$$ and $$B$$ are matrices

Last edited: Sep 13, 2010