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June 1st, 2019, 09:59 PM   #1
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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.
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June 2nd, 2019, 04:29 AM   #2
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Originally Posted by Integrator View Post
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.
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June 2nd, 2019, 04:48 AM   #3
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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.
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June 2nd, 2019, 05:44 AM   #4
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Originally Posted by Integrator View Post
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
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June 2nd, 2019, 06:36 AM   #5
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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.
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June 2nd, 2019, 12:34 PM   #6
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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
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June 2nd, 2019, 01:18 PM   #7
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$|i^x| = \left|\left(e^{i\pi/2}\right)^x\right| = |e^{i\pi x/2}| = 1$

$|x^i| =| e^{i\ln(x)}|$

$x > 0 \Rightarrow | e^{i\ln(x)}| = 1$

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....

$|x^i|+|i^x| = \begin{cases}1+e^{-\pi}&x\leq 0\\1&x=0\\2 &0<x\end{cases}$
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Last edited by romsek; June 2nd, 2019 at 01:21 PM.
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June 2nd, 2019, 08:44 PM   #8
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Quote:
Originally Posted by romsek View Post
$|i^x| = \left|\left(e^{i\pi/2}\right)^x\right| = |e^{i\pi x/2}| = 1$

$|x^i| =| e^{i\ln(x)}|$

$x > 0 \Rightarrow | e^{i\ln(x)}| = 1$

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....

$|x^i|+|i^x| = \begin{cases}1+e^{-\pi}&x\leq 0\\1&x=0\\2 &0<x\end{cases}$
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
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June 2nd, 2019, 09:45 PM   #9
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Quote:
Originally Posted by Integrator View Post
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
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June 2nd, 2019, 11:04 PM   #10
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Quote:
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good for you
OK , and for you how it is?

All the best,

Integrator
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