Previous | Next --- Slide 19 of 32
Back to Lecture Thumbnails
jzhanson

It took me a minute to grasp what Euler's identity $e^{i \pi} + 1 = 0$ meant in the context of complex "geometry"---it's just plugging in $\pi$ for $\theta$ in Euler's formula, which we now know corresponds to creating a rotation around the real/complex origin by $\pi$ radians, and since the coefficient of the right side $e^{i \theta}$ is 1, that means that we're rotating 1 by $\pi$ radians around the origin, giving us $-1$:

$$1 \cdot e^{i \pi} = \cos{\pi} + i \sin{\theta}$$ $$1 \cdot e^{i \pi} = -1 + i \cdot 0$$ $$1 \cdot e^{i \pi} = -1$$

And a simple manipulation gives us Euler's identity.

More interestingly, however, I came upon the brainteaser of what $i$ raised to the $i$-th power was---interestingly, it's a real number!

We begin by plugging in $\theta = \frac{\pi}{2}$ into Euler's formula:

$$e^{i \frac{\pi}{2}} = \cos{\frac{\pi}{2}} + i \sin{\frac{\pi}{2}}$$ $$e^{i \frac{\pi}{2}} = 0 + i \cdot 1 = i$$

We then raise both sides to the $i$-th power, giving $$e^{({i \frac{\pi}{2}})^i} = i^i$$

Simplifying the left side (and swapping sides), we arrive at $$i^i = e^{i \cdot i \frac{\pi}{2}} = e^{-\frac{\pi}{2}}$$

Even more interesting is that we can plug in any number of the form $\frac{\pi}{2} + n \cdot 2 \pi$ for any integer $$n$$ to arrive at a different value for $i^i$, but we consider the one where $n = 0$ to be the "principle" value.