Finding the solutions of a complex number

[EmmaK]

Homework Statement

Find the 3 solutions of e^i\pi/3z³=1/(1+i)

Homework Equations

e^i\theta=cos(\theta)+isin(\theta)

The Attempt at a Solution

i have put i/(1+i) into polar form,1/\sqrt{2} e^{i\stackrel{\pi}{4}}

So i get z³ = \stackrel{1}{\sqrt{2}}e^i-\pi/12

Then i got stuck... z³=r³e^i\theta

So shouldn't r³=1/\sqrt{2} and \theta=-\pi/36 ..but that's only 1 solution?

 

[latentcorpse]

well.
if z^3=\frac{1}{\sqrt{2}} e^{-\frac{i \pi}{12}}
then z=(\frac{1}{\sqrt{2}})^{\frac{1}{3}} e^{-\frac{i \pi}{36}}

 

[EmmaK]

but that is only one solution and it asks for 3

 

[Count Iblis]

You can add any multiple of 2 pi i to theta.

 

[EmmaK]

ahh, of course. thank you!

 

[RoyalCat]

ahh, of course. thank you!

Note that for unique solutions, you need to add n\cdot 2\pi i to the exponent of the complex number describing z^3

Otherwise you're just describing the same number over and over!

 

[EmmaK]

haha, oh yea.
where do you get n2/pi i from?

 

[RoyalCat]

haha, oh yea.
where do you get n2/pi i from?

That's how much you need to add to the angle of the exponent, r e^{i\theta} so that you get the same value.

re^{i\theta}=re^{i(\theta+2\pi)}

You can easily see this using Euler's identity since the sine and cosine both have a period of 2\pi radians.

When you take the cube root, you use De-Moivre and divide the angle by 3. Note that you get different angles depending on whether you add 2\pi once, twice, or three times to the original exponent's angle.

 

[latentcorpse]

yes i should have mentioned that. sorry.

 

[EmmaK]

so the final answer is \stackrel{n2\pi}{3}? i think i just misread your post as \stackrel{2n}{\pi} or something :)

 

[latentcorpse]

well, is (\frac{2n \pi}{3})^3=\frac{1}{\sqrt{2}}e^{-\frac{1 \pi}{12}}?

go to my first line of working in post 2, the other two solutions will correspond to e^{-\frac{25 i \pi}{12}} and e^{-\frac{49 i \pi}{12}}.
then of course, you have to do the division by 3 etc as before to get to the final answer.