#2**+10 **

Hi Michael and guest,

I thought it was interesting too.

Here is a proof.

You may like to watch this video on finding the arg(z) first. It is good.

but really |z| is just the distance Z is from (0,0) on the complex number plane

and

arg(z) is just the angle z makes with the positive real axis. (at the origin of course)

\(i^i=e^{-\pi/2}\\ proof\\ i^i=e^{[ln(i^i)]}\\ i^i=e^{i[ln(i)]}\\ \qquad\mbox{Now ln(z)=ln|z|+i*arg(z) so}\\ \qquad ln(i)=ln|i|+i*arg(i)\\ \qquad ln(i)=ln(1)+i*\frac{\pi}{2}\\ \qquad ln(i)=i*\frac{\pi}{2}\\ i^i=e^{i*i*\frac{\pi}{2}}\\ i^i=e^{-1*\frac{\pi}{2}}\\ i^i=e^{{\frac{-\pi}{2}}}\\ \)

.Melody Oct 17, 2015

#1**+5 **

You know what "i" stands for? Of course, it stands for √−1. So, what made you ask this question? Is it an assignment or just curiosity on your part? Or, are you studying "Complex numbers?" It's an interesting question, however!. But, I'm afraid that you may get lost in the explanation. It has a numerical value of: e^(-Pi/2)=0.207879576.........etc.

Guest Oct 17, 2015

#2**+10 **

Best Answer

Hi Michael and guest,

I thought it was interesting too.

Here is a proof.

You may like to watch this video on finding the arg(z) first. It is good.

but really |z| is just the distance Z is from (0,0) on the complex number plane

and

arg(z) is just the angle z makes with the positive real axis. (at the origin of course)

\(i^i=e^{-\pi/2}\\ proof\\ i^i=e^{[ln(i^i)]}\\ i^i=e^{i[ln(i)]}\\ \qquad\mbox{Now ln(z)=ln|z|+i*arg(z) so}\\ \qquad ln(i)=ln|i|+i*arg(i)\\ \qquad ln(i)=ln(1)+i*\frac{\pi}{2}\\ \qquad ln(i)=i*\frac{\pi}{2}\\ i^i=e^{i*i*\frac{\pi}{2}}\\ i^i=e^{-1*\frac{\pi}{2}}\\ i^i=e^{{\frac{-\pi}{2}}}\\ \)

Melody Oct 17, 2015

#3**+10 **

I've never seen this treated so I thought I'd take a deeper look.

\(c = r_c e^{\imath \theta_c} \\ c^\imath = \left( r_c e^{\imath \theta_c}\right)^\imath = r_c^\imath e^{\imath^2 \theta_c} = r_c^\imath e^{-\theta_c} \\ \mbox{Now what is }r_c^\imath? \\ r_c = e^{\ln(r_c)} \\ r_c^\imath = \left(e^{\ln(r_c)}\right)^\imath = e^{\imath \ln(r_c)} \\ \mbox{so }c^\imath = e^{\imath \ln(r_c)} e^{-\theta_c}= \\ e^{-\theta_c}\left(\cos(\ln(r_c))+\imath \sin(\ln(r_c))\right) \\ \mbox{Letting }c=\imath \\ r_c=1, \theta_c=\dfrac \pi 2 \\ \imath^\imath = e^{-\frac \pi 2} \left(\cos(\ln(1)) + \imath \sin(\ln(1))\right) = \\ e^{-\frac \pi 2}\left(\cos(0)+\imath \sin(0)\right)=e^{-\frac \pi 2} \\ \mbox{which is in agreement with Melody's answer}\)

.Rom Oct 18, 2015