Question

Let z=1- square root of 3i. Find z^7

Answer 1

\[\Large\bf\sf z\quad=\quad 1-\sqrt3\mathcal i\]Factoring out a 2 from each term,\[\Large\bf\sf z\quad=\quad 2\left(\frac{1}{2}-\frac{\sqrt3}{2}\mathcal i\right)\]We can write it in terms of sine and cosine from here.
Following along so far?

Answer 2

Yeah I got it so far

Answer 3

The imaginary component is negative, real is positive.. so that puts ussssss in theeeeeee uhhhhhh... 4th quadrant.

So our angle is 5pi/3, I think.

Answer 4

In exponential form we have:\[\Large\bf\sf z\quad=\quad 2 e^{\mathcal i 5\pi/3}\]

Answer 5

z^7 shouldn't be too bad from there.
Unless we're supposed to include all of the angles.. grr I always forget with these :p

Answer 6

Wait im sorry im still confused because my teacher doesn't do it that way... Where would I go from there?

Answer 7

From here,\[\Large\bf\sf z\quad=\quad 2\left(\frac{1}{2}-\frac{\sqrt3}{2}\mathcal i\right)\]Recognize that,\[\Large\bf\sf \cos\frac{5\pi}{3}\quad=\quad \frac{1}{2}\]\[\Large\bf\sf \sin\frac{5\pi}{3}\quad=\quad \frac{\sqrt3}{2}\]So we can write our expression as,\[\Large\bf\sf z\quad=\quad 2\left(\cos\frac{5\pi}{3}+\mathcal i \sin\frac{5\pi}{3}\right)\]That look familiar so far?

Answer 8

Have you learned about `De Moivre's Theorem`?

Answer 9

Woops sin(5pi/3) = -sqrt3/2.
Typo there. Shouldn't change anything though.

Answer 10

Oh okay yeah that's starting to look familiar

Answer 11

So then where would you go from there to get z^7?

Answer 12

So we have z.
We'll raise both sides to the 7th power.

Answer 13

The reason that's important is because you mustn't forget to put the power on the 2 as well.

Answer 14

\[\Large\bf\sf z^7\quad=\quad 2^7\left(\cos\frac{5\pi}{3}+\mathcal i \sin\frac{5\pi}{3}\right)^7\]Have you learned about `De Moivre's Theorem`?

Answer 15

Yeah we went over De Moivre's Theorem a little

Answer 16

De Moivre's will allow us to bring the 7 inside of the argument of our sine and cosine.

Answer 17

\[\Large\bf\sf z^7\quad=\quad 2^7\left(\cos\left[7\cdot\frac{5\pi}{3}\right]+\mathcal i \sin\left[7\cdot\frac{5\pi}{3}\right]\right)\]

Answer 18

So 128=(cos 35pi/3 + isin 35pi/3)?

Answer 19

Oh oh oh I screwed up on the angle I think.
Pretty sure we want to use the `principle angle` which would be from -pi to pi.
So for our angle we probably want to use -pi/3 instead of 5pi/3

Answer 20

But uhhh yahhhh I think that puts us on the right track :U

Answer 21

oh okay so 128= (cos 7pi/3+ isin 7pi/3)?

Answer 22

Why is there an equals sign in there? +_+

Answer 23

because 2^7 = 128

Answer 24

oh oops nevermind haha I get it

Answer 25

Ya something like that I think,\[\Large\bf\sf z^7\quad=\quad 128\left(\cos\left[-\frac{7\pi}{3}\right]+\mathcal i \sin\left[-\frac{7\pi}{3}\right]\right)\]negative on the angles.
But then uhhh we can relize that umm....

Answer 26

\[\Large\bf\sf -\frac{7\pi}{3}\quad=\quad -\frac{\pi}{3}-2\pi\]Since sine and cosine are periodic in 2pi, we can rewrite our angle as,\[\Large\bf\sf -\frac{\pi}{3}\]

Answer 27

\[\Large\bf\sf z^7\quad=\quad 128\left(\cos\left[-\frac{\pi}{3}\right]+\mathcal i \sin\left[-\frac{\pi}{3}\right]\right)\]

Answer 28

And then we can write our cosine and sine back in terms of their reference values.
The same ones from before.\[\Large\bf\sf z^7\quad=\quad 128\left(\frac{1}{2}- \frac{\sqrt3}{2}\mathcal i\right)\]

Answer 29

Then put a 2 back into the brackets.

Answer 30

We kinda went in circles but it really just depends on what form you need your answer in.

Answer 31

Oh so the answer would be \[128(1/2\left[ 2 \right]- \sqrt{3}/2i \left[ 2 \right])\]