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In the complex plane there is a nice relationship between rectangular, polar, and exponential coordinates:

$$(x+iy) = r(\cos\theta + i~\sin\theta) = re^{i\theta} $$

$$where~~x ,y ,\theta, r \in \unicode{x211D}$$

The three relationships above are all I ever hear about when I read about Complex numbers. However after playing around with imaginary number of various real powers $i^0, i^1, i^2,...$ it becomes clear imaginary powers are cyclical. Thinking about it more it become clear that imaginary powers are not just cyclical but points on a circle. From here it is easy to figure out that $re^{i\theta} = ri^{2\theta/\pi}$. Thus we now have four relationships.

$$ (x+iy) = r(cos\theta + i~sin\theta) = re^{i\theta} = ri^{2\theta/\pi} $$

$$where~~x ,y ,\theta, r \in \unicode{x211D}$$

My question is why do people never mention this fourth relationship? For instance there is a very clear symmetry between $re^{i\theta} = ri^{2\theta/\pi}$. In both of these we can represent any complex number with $r$ (radius) and $\theta$ (angle). However in the former we are representing an arbitrary complex number as a real number raised to an imaginary power $re^{i\theta}$, and in the latter we are representing an arbitrary complex number as an imaginary number raised to a real power $ri^{2\theta/\pi}$!

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  • $\begingroup$ Because $x^y$ is not very well-behaved, and $i^y$ is tricky. $x^y$ is only single-valued if $x=e$ or $y$ is an integer... $\endgroup$ – Thomas Andrews Feb 18 '15 at 1:34
  • $\begingroup$ Basically, you can't assume that $(x^y)^z=x^{yz}$ for all complex $y,z$. $\endgroup$ – Thomas Andrews Feb 18 '15 at 1:36
  • $\begingroup$ Alternatively, you can see this as just a change of units, where there are $2\pi$ radians in a circle, but $4$ "new unit" in this new form. So it's not particularly new - indeed, if you use degrees as your unit, then it can be seen in this way as $r\zeta^\theta$ whre $\zeta=e^{\pi i/180}$. $\endgroup$ – Thomas Andrews Feb 18 '15 at 1:39
  • $\begingroup$ The formulas I gave above are well-formed because $x, y, r, \theta, \in \unicode{x211D}$ $\endgroup$ – Soto Feb 18 '15 at 16:05
  • $\begingroup$ $i^x$ is not well-defined, however. Nobody said anything about well-formed. $\endgroup$ – Thomas Andrews Feb 18 '15 at 16:14

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