Question

Hello can you help me find the complex cube roots of 8(cos(4pi/5)+isin(4pi/5))

Answer 1

\[\Large\rm z=8\left[\cos\left(\frac{4\pi}{5}\right)+\mathcal i \sin\left(\frac{4\pi}{5}\right)\right]\]\[\Large\rm z^{1/3}=8^{1/3}\left[\cos\left(\frac{4\pi}{5}\right)+\mathcal i \sin\left(\frac{4\pi}{5}\right)\right]^{1/3}\]

Answer 2

Hey There Diana :)
Do you remember De'Moivre's Theorem?

Answer 3

Alright, I'll leave some notes in case you come back.
Before we apply De'Moivre's Theorem, let's rewrite our angles inside of our sine and cosine, we need to allow for rotation.\[\rm z^{1/3}=8^{1/3}\left[\cos\left(\frac{4\pi}{5}+2k \pi\right)+\mathcal i \sin\left(\frac{4\pi}{5}2k \pi\right)\right]^{1/3}, \quad k=0,\pm1,\pm2,\pm3,...\]We can add any number of full spins.
We're allowing our angle to be 4pi/5 or 14pi/5 or 24pi/5 and so on.... because when we take the sine and cosine of those angles, they give us the same values that 4pi/5 would give.

Since this is the `third root`, we only want the first three positive values for k, starting from 0, after that our solutions will start to repeat.

Answer 4

\[\rm z^{1/3}=8^{1/3}\left[\cos\left(\frac{4\pi}{5}+2k \pi\right)+\mathcal i \sin\left(\frac{4\pi}{5}2k \pi\right)\right]^{1/3}, \quad k=0,1,2\]Applying De'Moivre's Theorem lets us bring the exponent in as a factor on the angle.\[\rm z^{1/3}=8^{1/3}\left[\cos\left(\frac{4\pi}{3\cdot5}+\frac{2k \pi}{3}\right)+\mathcal i \sin\left(\frac{4\pi}{3\cdot5}+\frac{2k \pi}{3}\right)\right], \quad k=0,1,2\]

Answer 5

Simplify,
then plug in your k's to get the three results :)

Answer 6

I'll show you the first one just so it's clear.
Err I'll show you k=2 actually.

Answer 7

\[\rm z_2^{1/3}=8^{1/3}\left[\cos\left(\frac{4\pi}{3\cdot5}+\frac{2\cdot2 \pi}{3}\right)+\mathcal i \sin\left(\frac{4\pi}{3\cdot5}+\frac{2\cdot2 \pi}{3}\right)\right]\]\[\rm z_2^{1/3}=2\left[\cos\left(\frac{4\pi}{15}+\frac{20\pi}{15}\right)+\mathcal i \sin\left(\frac{4\pi}{15}+\frac{20\pi}{15}\right)\right]\]\[\rm z_2^{1/3}=2\left[\cos\left(\frac{24\pi}{15}\right)+\mathcal i \sin\left(\frac{24\pi}{15}\right)\right]\]\[\rm z_2^{1/3}=2\left[\cos\left(\frac{8\pi}{5}\right)+\mathcal i \sin\left(\frac{8\pi}{5}\right)\right]\]Mmm that doesn't work out so nice -_- hmm
probably need a calculator from here.\[\Large\rm \rm z_2^{1/3}=2\left[0.309-\mathcal i 0.951\right]\]\[\Large\rm \rm z_2^{1/3}=0.6180-1.902\mathcal i\]

Answer 8

So that would be one of our solutions :U

Answer 9

Thank you so much!

Answer 10

Did you find the other roots bunny? c:

Answer 11

not really

Answer 12

the process too confusing? :o

Answer 13

ha

Answer 14

yeah

Answer 15

hmm :d

Answer 16

@zepdrix Can you help me find the other two please.

Answer 17

When k=0, it's pretty straight forward.
It's the equation you started with,\[\Large\rm z^{1/3}=8^{1/3}\left[\cos\left(\frac{4\pi}{5}\right)+\mathcal i \sin\left(\frac{4\pi}{5}\right)\right]^{\color{orangered}{1/3}}\]no rotation.
And apply De'Moivre's Theorem,\[\Large\rm z^{1/3}=8^{1/3}\left[\cos\left(\color{orangered}{\frac{1}{3}}\cdot\frac{4\pi}{5}\right)+\mathcal i \sin\left(\color{orangered}{\frac{1}{3}}\cdot\frac{4\pi}{5}\right)\right]\]And let your calculator do the rest :o

Answer 18

When k=1, we'll add 1 full spin to our angle.\[\Large\rm z^{1/3}=8^{1/3}\left[\cos\left(\frac{4\pi}{5}+2\pi\right)+\mathcal i \sin\left(\frac{4\pi}{5}+2\pi\right)\right]^{\color{orangered}{1/3}}\]Applying De'Moivre,\[\large\rm z^{1/3}=8^{1/3}\left[\cos\left(\color{orangered}{\frac{1}{3}\cdot}\frac{4\pi}{5}+\color{orangered}{\frac{1}{3}\cdot}2\pi\right)+\mathcal i \sin\left(\color{orangered}{\frac{1}{3}\cdot}\frac{4\pi}{5}+\color{orangered}{\frac{1}{3}\cdot}2\pi\right)\right]\]Simplify, and calculator :O

Answer 19

how would I put that in my calculator

Answer 20

@zepdrix I think I got it k 0=2.0004+0.0292i and k 1=1.9972+0.102i