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 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
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 Originally Posted by Integrator Hello all, Calculate $\displaystyle |x^i|+|i^x|$ where $\displaystyle x\in \mathbb R$ and $\displaystyle i^2=-1$. All the best, Integrator
I don't have a full answer for you, but
$\displaystyle |i^x| = \left | \left ( e^{i \pi / 2} \right )^x \right | = \left | e^{ix \pi / 2} \right | = 1$

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. W|A 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
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 Originally Posted by topsquark I don't have a full answer for you, but $\displaystyle |i^x| = \left | \left ( e^{i \pi / 2} \right )^x \right | = \left | e^{ix \pi / 2} \right | = 1$ 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. W|A agrees but doesn't provide a derivation. Note that there is a problem with x = 0... arg(0) is not defined. -Dan
Hello,

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
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 Originally Posted by Integrator Hello, 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
Well, if I got the |x^i| right then |x^i| + |i^x| = 1 + 1 = 2.

-Dan June 2nd, 2019, 06:36 AM   #5
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 Originally Posted by topsquark Well, if I got the |x^i| right then |x^i| + |i^x| = 1 + 1 = 2. -Dan
Your answer is valid for any $\displaystyle x\in \mathbb R$?If $\displaystyle x=-1$ , then $\displaystyle |x^i|+|i^x|=?$
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
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 Originally Posted by Integrator Your answer is valid for any $\displaystyle x\in \mathbb R$?If $\displaystyle x=-1$ , then $\displaystyle |x^i|+|i^x|=?$ Thank you very much! All the best, Integrator
It's still 2. (But like I said someone needs to double-check my formula.)

-Dan 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 June 2nd, 2019, 08:44 PM #8 Member Joined: Aug 2018 From: România Posts: 88 Thanks: 6 Quote:  Originally Posted by romsek $|i^x| = \left|\left(e^{i\pi/2}\right)^x\right| = |e^{i\pi x/2}| = 1|x^i| =| e^{i\ln(x)}|x > 0 \Rightarrow | e^{i\ln(x)}| = 1$For$x < 0\ln(x) = \ln(-|x|) = \ln(-1)+\ln(|x|) = i \pi + \ln(|x|)\exp\left(i(i\pi + \ln(|x|))\right) = \exp(-\pi + i\ln(|x|)) = e^{-\pi}e^{i\ln(|x|)}|x^i| =|e^{-\pi}e^{i\ln(|x|)}|=|e^{-\pi}||e^{i\ln(|x|}|= |e^{-\pi}| = e^{-\pi},~x< 0x=0 \Rightarrow |x^i| = 0^i = 0x=0 \Rightarrow |x^i|=0$so....$|x^i|+|i^x| = \begin{cases}1+e^{-\pi}&x\leq 0\\1&x=0\\2 &0
Hello,

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
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Quote:
 Originally Posted by Integrator Hello, 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
good for you June 2nd, 2019, 11:04 PM   #10
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 Originally Posted by romsek good for you
OK , and for you how it is?

All the best,

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