I’m asked to find all the values of iiiii^i and then plot them in the complex plane.

First I used Euler’s identity to show that

i=±ei2πn±πi/2i=±ei2πn±πi/2i = \pm e^{i2\pi n\pm \pi i/2},

so then

ii=±e−2πn∓π/2ii=±e−2πn∓π/2i^i = \pm e^{-2\pi n \mp \pi/2}

Now that I have this, I’m not sure how to plot it. Does it go in the complex z plane of the complex w plane?

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There no ±in the argument of iii.

– Bernard

Oct 21 at 0:24

In full generality you don’t have an identity (−x)y=−(xy)(-x)^y = -(x^y). But having that toggled sign is making things more complicated than they need to be.

– Hurkyl

Oct 21 at 0:24

math.stackexchange.com/questions/301929/…

– rlartiga

Oct 21 at 0:25

Okay, so if it were just ii=e−2πn−π/2i^i = e^{-2\pi n-\pi /2} how would I got about plotting it?

– Spuds

Oct 21 at 0:26

1

@SSimple Art: in a very weak sense â€” per chance (as though two computational errors yielded the right answer). It lacks rigour, from my point of view.

– Bernard

Oct 21 at 1:13

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1 Answer

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In complex numbers you have to take care with power functions.

In complex analysis you have infinity logarithms. A logarithm function ll in a region D is a holomorphic function in DD such that exp(l(z))=z\exp(l(z))=z for all z∈Dz \in D. As exp\exp is not injective in C\mathbb{C}, a complex number can have infinite logarithms. But you also know that the period of complex exponential function is 2πiZ2\pi i \mathbb{Z}, so if ˉl\bar l is a logarithm function in DD, then other logarithms functions in DD are given by l=ˉl+2πikl = \bar l + 2 \pi i k for some k∈Zk \in \mathbb{Z}.

The principal branch of logarithm is defined in C\mathbb{C} without the real negative axis, by l(z)=log|z|+iθl(z)= \log|z|+i \theta, where θ\theta is the principal argument of zz, and log\log is the usual real logarithm.

Now that you fix a logarithm function ll, you can define zσ=exp(σl(z))z^\sigma = \exp(\sigma l(z)). With the principal branch, l(i)=log|1|+iπ2=iπ2l(i)=\log|1|+i\frac{\pi}{2}=i\frac{\pi}{2}, so ii=exp(il(i))=exp(i2π2)=exp(−π2)i^i = \exp(i l(i))=\exp(i^2\frac{\pi}{2})=\exp(-\frac{\pi}{2}).

So all values of iii^i are given by exp(−π2+2πik)\exp(-\frac{\pi}{2}+2\pi ik) with k∈Zk \in \mathbb{Z}.

Notice that exp(2πik)=1\exp(2\pi ik)=1 for all k∈Zk \in \mathbb{Z} so your plot its just one point, exp(−π2).\exp(-\frac{\pi}{2}).