## arindameducationusc one year ago Can anyone explain x^pi?

1. ParthKohli

It's $$x$$ raised to the power $$\pi$$.

2. arindameducationusc

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

3. arindameducationusc

Ya this one

4. arindameducationusc

The imaginary part and real part...

5. arindameducationusc

Can you explain...

6. ParthKohli

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

7. ParthKohli

You do understand complex numbers, right?

8. arindameducationusc

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

9. ParthKohli

Yeah. For negative $$x$$, this function has complex values.

10. arindameducationusc

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

11. Astrophysics
12. Astrophysics

khanacademy is awesome for all math

13. Astrophysics

as well

14. arindameducationusc

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

15. anonymous

Its just x^3.14

16. 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.

17. arindameducationusc

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

18. 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

19. Astrophysics

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