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In our last post, we discussed raising a real number to a complex power.

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We have defined $e^z$ (therefore $n^z$), however we now want to define $z^n$, where z is a complex number.
$$z^n = e^{n\log(z)}$$
However, we immediately run into a problem. We have not defined complex logarithms.

Complex Logarithms

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We begin by writing the equation
$$w = \log(z) \to e^w = z$$
We can rewrite w as $u + iv$ and $z$ as $re^{i\theta}$
$$e^{u + iv} = re^{i\theta}$$
So, $e^u = r$ and $e^{iv} = e^{i\theta}$
We now can express u and v.
$$u = \ln(r), v = \theta + 2k\pi $$
Now, we combine our answer to get the logarithm.
$$w = \ln(r) + i\left(\theta + 2k\pi\right)$$

Quick Example

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$$\ln(-1)$$
Well, we have to figure out two things: the modulus (magnitude) of this complex number and its argument (angle).
The magnitude is $1$, and the angle is $\pi$.
So, the logarithm equals
$$\ln(1) + i\left(\pi + 2k\pi\right)$$
$$i\left(\pi + 2k\pi\right)$$

So, what is i^i?

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$$i^i = e^{i\ln(i)}$$
$$e^{i\ln(i)} = e^{i*\left(i\frac{\pi}{2}\right)}$$
$$= e^{-\frac{\pi}{2}} = \boxed{0.207}$$
That is a pretty crazy result, because we raised a complex number to a complex power and got a real number. Another way to think about this is
$$i = e^{i\frac{\pi}{2}} \to i^i = e^{i\frac{\pi}{2} \cdot i} $$
Now, we get the same result, $e^{-\frac{\pi}{2}}$