Raising e^(3πi/4) to the 11th power in cartesian form

[charmedbeauty]

Homework Statement

α=2e^3∏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 e^3∏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.

 

[HallsofIvy]

Oh, dear! I just divided wrong! 33/7= 8+ 1/7.

 

[clamtrox]

so if I let k=15

I get e^3∏i/4

I get something different.. $(e^{3 \pi i /4} )^{11} = e^{33 \pi i /4} = e^{8 \pi i + \pi i/4} = e^{\pi i/4}$

 

[Infinitum]

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

 

[charmedbeauty]

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.