Solving (e[SUP]i[/SUP])[SUP]i[/SUP]

[Icetray]

Hey guys,

I have the following question that I am trying to solve. Would really like to solve it on my own first so I'd appreciate hints first! :D

Homework Statement

Solve (eⁱ)ⁱ

The attempt at a solution

(eⁱ)ⁱ = (cos pi/pi + isin pi/pi)ⁱ
= (cos i + isin i) (Euler's Formula)

I'm not sure if I'm even right till this point but this is the furthest that I have managed to come so far. ):

Thanks in advance guys!

 

[HallsofIvy]

It might be more helpful to note that if y= (e^i)^i then ln(y)= i ln(e^i)= i^2= -1. And, of course, ln(re^{i\theta}= ln(r)+ i(\theta+ 2k\pi).

 

[Icetray]

It might be more helpful to note that if y= (e^i)^i then ln(y)= i ln(e^i)= i^2= -1. And, of course, ln(re^{i\theta}= ln(r)+ i(\theta+ 2k\pi).

Wouldn't that leave me with ln y = -1 and ln y = ln i?

Anyway thanks for your quick reply HallsofIvy! Your quick reply is very much appriciated as usual! :D

 

[NascentOxygen]

Solve (eⁱ)ⁱ

(eⁱ)ⁱ = e^i*i = e^-1 = ...

 

[Icetray]

(eⁱ)ⁱ = e^i*i = e^-1 = ...

I thought of this but didn't think it would be this simple hahaha. So just the calculator value of e^-1? :D

 

[HallsofIvy]

Wouldn't that leave me with ln y = -1 and ln y = ln i?

No, "2k\pi is always a multiple of 2\pi, not \pi. You would only get ln y= -1.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Anyway thanks for your quick reply HallsofIvy! Your quick reply is very much appriciated as usual! :D </div> </div> </blockquote> <br /> <br /> <blockquote data-attributes="" data-quote="NascentOxygen" data-source="post: 3461989" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> NascentOxygen said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> (eⁱ)ⁱ = e^i*i = e^-1 = ... </div> </div> </blockquote> Ouch! Of course!

 

[dynamicsolo]

I had a post here which I deleted, as I missed the same thing. HallsofIvy's point is correct in general, in that complex numbers can have an infinite number of complex logarithms (think of what happens with taking roots with DeMoivre's Theorem, where there are six sixth roots of 1 , and so forth; it gets worse with logarithms).

However, they were cute with this problem because all of the logarithms fall in the same spot on an Argand diagram.

If we work with the complex exponential forms for sine and cosine, e.g., \cos z = \frac{e^{iz} + e^{-iz}}{2}, we just get one real value for ( cos i + i sin i ) . Other staged exponentials involving complex numbers may not resolve so neatly...