Little confused about raising a complex number to a power.

charmedbeauty
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Homework Statement



α=2e3∏i/4

find α11 in cartesian form.


Homework Equations





The Attempt at a Solution



It's been a while since I've done these but from what remember you add 2kpi to get exp in the range of -∏,∏.

so if I let k=15

I get e3∏i/4

but the sltn says it needs to be raised to ∏i/4

can some one please tell me why, should I be adding 4k∏ since it is divided by 4?

Thanks.
 
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Oh, dear! I just divided wrong! 33/7= 8+ 1/7.
 
Last edited by a moderator:
charmedbeauty said:
so if I let k=15

I get e3∏i/4

I get something different.. (e^{3 \pi i /4} )^{11} = e^{33 \pi i /4} = e^{8 \pi i + \pi i/4} = e^{\pi i/4}
 
HallsofIvy said:
What you have is correct. The exponent is 3\pi i/4, not \pi i/4.

Hmm, I get the exponent as \pi i/4.

We have,

\vec{p} = e^{3\pi i/4}

Raising the power to 11,

\vec{t} = e^{33\pi i/4}

Looking at 33\pi /4 we see that it crosses the first quadrant(2n\pi), 4 times, so that gives an angular displacement of 8\pi. Let x be the angle in cartesian range we are looking for,

8\pi + x = \frac{33\pi}{4}Edit : Just saw your edit :-p
 
HallsofIvy said:
Oh, dear! I just divided wrong! 33/7= 8+ 1/7.

oops I did the same thing, oh damn!

Thanks for clearing that up anyhow.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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