arindameducationusc
  • arindameducationusc
Can anyone explain x^pi?
Mathematics
chestercat
  • chestercat
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ParthKohli
  • ParthKohli
It's \(x\) raised to the power \(\pi\).
arindameducationusc
  • arindameducationusc
yes, i will upload the graph... I have a doubt. wait..
arindameducationusc
  • arindameducationusc
Ya this one
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arindameducationusc
  • arindameducationusc
The imaginary part and real part...
arindameducationusc
  • arindameducationusc
Can you explain...
ParthKohli
  • ParthKohli
For a positive number, this is of course a purely real number. Thus the imaginary part is zero.
ParthKohli
  • ParthKohli
You do understand complex numbers, right?
arindameducationusc
  • arindameducationusc
Yes I understand Complex numbers. So that means we can graph complex numbers? I didn't know that!
ParthKohli
  • ParthKohli
Yeah. For negative \(x\), this function has complex values.
arindameducationusc
  • arindameducationusc
Do you have any reference in which I can study complex number graphs, any good books or video link?
Astrophysics
  • Astrophysics
http://www.mathsisfun.com/numbers/complex-numbers.html
Astrophysics
  • Astrophysics
khanacademy is awesome for all math
Astrophysics
  • Astrophysics
as well
arindameducationusc
  • arindameducationusc
Then if x=3, then 3^3.14 should be real, but in graph it is imaginary..
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anonymous
  • anonymous
Its just x^3.14
zepdrix
  • zepdrix
@arindameducationusc You'll notice that the orange line stays on the x-axis when x is positive. The imaginary part is zero for all positive x. So there is no imaginary part over there.
arindameducationusc
  • arindameducationusc
I got it...@zepdrix @jcoury @Astrophysics @ParthKohli Thank you to all for helping.....
phi
  • phi
you may know \[ e^{i\pi}= \cos\pi + i \sin \pi = -1 + 0 \ i = -1 \] when x is negative, x^pi is the same as \[ (- |x|)^\pi = \left( e^{i\pi} |x|\right)^\pi \\= e^{i\pi^2} |x|^\pi= (\cos\pi^2 + i \sin \pi^2)|x|^\pi \\ (-x)^\pi \cos\pi^2 + i (-x)^\pi \sin \pi^2 \] which has a non-zero imaginary component
Astrophysics
  • Astrophysics
^ Yes, that's very useful, Euler's equation

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