# raising complex numbers to complex exponents?

• Mar 25th 2011, 08:14 PM
jamix
raising complex numbers to complex exponents?
Is it possible to calculate something like $(a + bi)^{c+di}$?

I recall reading somewhere that $i^i$ is calculable (I think its a real number), so I was pondering whether or not this was true for complex numbers in general?
• Mar 25th 2011, 08:20 PM
dwsmith
Quote:

Originally Posted by jamix
Is it possible to calculate something like $(a + bi)^{c+di}$?

I recall reading somewhere that $i^i$ is calculable (I think its a real number), so I was pondering whether or not this was true for complex numbers in general?

Yes.

$z^{\alpha}=e^{\alpha\ln(z)}$
• Mar 25th 2011, 08:25 PM
dwsmith
• Mar 25th 2011, 08:29 PM
jamix
Ah I think I just figured it out...

First you put the bottom number into polar coordinates $Re^{i0}$. Then you break up the exponent so we have the following:

$(Re^{i0})^c \cdot (Re^{i0})^{di}$ = $R \cdot (cos(0c) + isin(0c)) \cdot ( Re^{-d0})$ = $R^2 \cdot e^{-d0} \cdot ( cos(0c) + isin(0c))$
• Mar 25th 2011, 08:42 PM
Prove It
Yes you can. It's no simple task, but it requries you write the complex number in its exponential-polar form.

Complex Exponentiation -- from Wolfram MathWorld
• Mar 25th 2011, 09:05 PM
jamix
The Wolfram link really helped me understand what was being said in the other thread (which I couldn't follow).
• Mar 25th 2011, 09:06 PM
dwsmith
Quote:

Originally Posted by jamix
The Wolfram link really helped me understand what was being said in the other thread (which I couldn't follow).

I think it is easier to use the e^{aLn(z)}
• Mar 25th 2011, 09:23 PM
Prove It
Quote:

Originally Posted by dwsmith
I think it is easier to use the e^{aLn(z)}

There's no such thing as $\displaystyle \ln{z}$, you can only take a natural logarithm of positive real numbers.

However, $\displaystyle \log{z} = \ln{|z|} + i\arg{z}$ is acceptable. It is also needed to be able to write the answer explicitly in terms of its real and imaginary parts.
• Mar 25th 2011, 09:24 PM
dwsmith
Quote:

Originally Posted by Prove It
There's no such thing as $\displaystyle \ln{z}$, you can only take a natural logarithm of positive real numbers.

However, $\displaystyle \log{z} = \ln{|z|} + i\arg{z}$ is acceptable. It is also needed to be able to write the answer explicitly in terms of its real and imaginary parts.

$\log(z)=\log_e(z)=\text{Ln}(z)$
• Mar 25th 2011, 09:30 PM
dwsmith
• Mar 25th 2011, 10:02 PM
CaptainBlack
Quote:

Originally Posted by Prove It
There's no such thing as $\displaystyle \ln{z}$, you can only take a natural logarithm of positive real numbers.

However, $\displaystyle \log{z} = \ln{|z|} + i\arg{z}$ is acceptable. It is also needed to be able to write the answer explicitly in terms of its real and imaginary parts.

Rubbish, you just did or rather discovered its extension (note it is multi-valued, and we will need that if we are going to use it to take powers of complex numbers).

CB