raising complex numbers to complex exponents?

**jamix** · March 25th 2011, 08:14 PM

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?

**dwsmith** · March 25th 2011, 08:20 PM

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.

$z^{\alpha}=e^{\alpha\ln(z)}$

**dwsmith** · March 25th 2011, 08:25 PM

http://www.mathhelpforum.com/math-he...rs-174683.html

Go here

**jamix** · March 25th 2011, 08:29 PM

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:

$(Re^{i0})^c \cdot (Re^{i0})^{di}$ = $R \cdot (cos(0c) + isin(0c)) \cdot ( Re^{-d0})$ = $R^2 \cdot e^{-d0} \cdot ( cos(0c) + isin(0c))$

**Prove It** · March 25th 2011, 08:42 PM

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

**jamix** · March 25th 2011, 09:05 PM

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

**dwsmith** · March 25th 2011, 09:06 PM

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)}

**Prove It** · March 25th 2011, 09:23 PM

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.

**dwsmith** · March 25th 2011, 09:24 PM

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.

$\log(z)=\log_e(z)=\text{Ln}(z)$

**dwsmith** · March 25th 2011, 09:30 PM

Natural Logarithm -- from Wolfram MathWorld

Middle way down.

**CaptainBlack** · March 25th 2011, 10:02 PM

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

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