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Mathematics 10 Online
OpenStudy (arindameducationusc):

Can anyone explain x^pi?

Parth (parthkohli):

It's \(x\) raised to the power \(\pi\).

OpenStudy (arindameducationusc):

yes, i will upload the graph... I have a doubt. wait..

OpenStudy (arindameducationusc):

Ya this one

OpenStudy (arindameducationusc):

The imaginary part and real part...

OpenStudy (arindameducationusc):

Can you explain...

Parth (parthkohli):

For a positive number, this is of course a purely real number. Thus the imaginary part is zero.

Parth (parthkohli):

You do understand complex numbers, right?

OpenStudy (arindameducationusc):

Yes I understand Complex numbers. So that means we can graph complex numbers? I didn't know that!

Parth (parthkohli):

Yeah. For negative \(x\), this function has complex values.

OpenStudy (arindameducationusc):

Do you have any reference in which I can study complex number graphs, any good books or video link?

OpenStudy (astrophysics):

http://www.mathsisfun.com/numbers/complex-numbers.html

OpenStudy (astrophysics):

khanacademy is awesome for all math

OpenStudy (astrophysics):

as well

OpenStudy (arindameducationusc):

Then if x=3, then 3^3.14 should be real, but in graph it is imaginary..

OpenStudy (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.

OpenStudy (arindameducationusc):

I got it...@zepdrix @jcoury @Astrophysics @ParthKohli Thank you to all for helping.....

OpenStudy (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

OpenStudy (astrophysics):

^ Yes, that's very useful, Euler's equation

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