June 1st, 2019, 09:59 PM  #1 
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  A calculation
Hello all, Calculate $\displaystyle x^i+i^x$ where $\displaystyle x\in \mathbb R$ and $\displaystyle i^2=1$. All the best, Integrator Last edited by skipjack; June 1st, 2019 at 11:02 PM. 
June 2nd, 2019, 04:29 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle i^x = \left  \left ( e^{i \pi / 2} \right )^x \right  = \left  e^{ix \pi / 2} \right  = 1$ I believe that $\displaystyle \left  x^i \right  = e^{arg(x)} = 1$, given that x is real. That's what I recall but someone is going to have to check that for me. WA agrees but doesn't provide a derivation. Note that there is a problem with x = 0... arg(0) is not defined. Dan Last edited by skipjack; June 2nd, 2019 at 02:13 PM.  
June 2nd, 2019, 04:48 AM  #3  
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  Quote:
It is obvious that for $\displaystyle x=0$ we can not calculate... What values can take that amount for $\displaystyle x\neq 0$? Thank you very much! All the best! Integrator Last edited by skipjack; June 2nd, 2019 at 02:26 PM.  
June 2nd, 2019, 05:44 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  
June 2nd, 2019, 06:36 AM  #5  
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  Quote:
Thank you very much! All the best, Integrator Last edited by Integrator; June 2nd, 2019 at 06:39 AM.  
June 2nd, 2019, 12:34 PM  #6 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,266 Thanks: 934 Math Focus: Wibbly wobbly timeywimey stuff.  
June 2nd, 2019, 01:18 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 
$i^x = \left\left(e^{i\pi/2}\right)^x\right = e^{i\pi x/2} = 1$ $x^i = e^{i\ln(x)}$ $x > 0 \Rightarrow  e^{i\ln(x)} = 1$ For $x < 0$ $\ln(x) = \ln(x) = \ln(1)+\ln(x) = i \pi + \ln(x)$ $\exp\left(i(i\pi + \ln(x))\right) = \exp(\pi + i\ln(x)) = e^{\pi}e^{i\ln(x)}$ $x^i =e^{\pi}e^{i\ln(x)}=e^{\pi}e^{i\ln(x}= e^{\pi} = e^{\pi},~x< 0$ $x=0 \Rightarrow x^i = 0^i = 0$ $x=0 \Rightarrow x^i=0$ so.... $x^i+i^x = \begin{cases}1+e^{\pi}&x\leq 0\\1&x=0\\2 &0<x\end{cases}$ Last edited by romsek; June 2nd, 2019 at 01:21 PM. 
June 2nd, 2019, 08:44 PM  #8  
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  Quote:
No offense , but I do not understand!$\displaystyle 0^i$ is defined?What value has $\displaystyle 0^i$?I say that calcul of the sum $\displaystyle x^i+i^x$ can not be done for $\displaystyle x=0$... Thank you very much! All the best, Integrator  
June 2nd, 2019, 09:45 PM  #9 
Senior Member Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399  good for you

June 2nd, 2019, 11:04 PM  #10 
Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6  

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