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What is imaginary I raised to the power imaginary i?

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872
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What is I to the power I?

Oct 17, 2015
edited by Guest  Oct 17, 2015

#2
+109524
+10

Hi Michael and guest,

I thought it was interesting too.

Here is a proof.

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)

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

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Oct 17, 2015
edited by 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.

Oct 17, 2015
#2
+109524
+10

Hi Michael and guest,

I thought it was interesting too.

Here is a proof.

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
edited by Melody  Oct 17, 2015
#3
+6187
+10

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

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

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Oct 18, 2015