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gmandelbraat · Jun 27, 2011, 07:07 AM

How to compute for example 2^(2-i)?

ebaines · Jun 27, 2011, 07:21 AM

Start here:

$<br /> 2^{(2-i)} = \frac {2^2}{2^i}<br />$

Now from Euler's Formula: $e^{i \theta} \ =\ \cos \theta + i \sin \theta$ , and since $2 = e^{\ln 2}$ :

$<br /> 2^{i} = e^{(\ln 2) i} = \cos(\ln 2) + i \sin (\ln 2)<br />$

You can finish from here.

gmandelbraat · Jun 27, 2011, 07:49 AM

Thanks ebaines! It makes sense.

But what if the base is complex also? How to find the logarithm of a complex number?

jcaron2 · Jun 27, 2011, 08:17 AM

Unfortunately complex numbers don't have a well-defined log. Since sine and cosine are periodic, that means that $e^x=e^{x+2\pi i}=e^{x+2n\pi i}$ (where n is an integer).

Since the logarithm function is essentially answering the question "To what power do I need to raise the base to achieve the argument of the log?", you can imagine that there are actually an infinite number of answers. For example, let's take ln(i). Looking at Euler's formula, it should be self-evident that if you set $\theta=\frac{\pi}{2}$ , the right side of the equals sign comes out to i. Thus, it stands to reason that ln(i)= $\frac{\pi}{2}i$ . However, it also works if $\theta=\frac{5\pi}{2}i$ or $\theta=\frac{9\pi}{2}i$ , etc.

Thus, if you try to take a complex number to a complex power (for example i^i), there are actually an infinite number of answers. I think it's normally acceptable, however, to give only the answer corresponding to the logarithm with the smallest magnitude (similar to the way the arctan function on a calculator usually only provides solutions in the first and fourth quadrants, even though there are technically an infinite number of answers).

$i^i=e^{\ln(i) \cdot i}=e^{\frac{\pi}{2}i \cdot i}= e^{-\frac{\pi}{2}} \approx 0.2079$