Finding the solutions of a complex number

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
10 replies · 2K views
EmmaK
Messages
24
Reaction score
0

Homework Statement



Find the 3 solutions of ei[tex]\pi[/tex]/3z3=1/(1+i)

Homework Equations


ei[tex]\theta[/tex]=cos([tex]\theta[/tex])+isin([tex]\theta[/tex])


The Attempt at a Solution



i have put i/(1+i) into polar form,1/[tex]\sqrt{2}[/tex] ei[tex]\stackrel{\pi}{4}[/tex]

So i get z3 = [tex]\stackrel{1}{\sqrt{2}}[/tex]ei-[tex]\pi/12[/tex]

Then i got stuck... z3=r3ei[tex]\theta[/tex]

So shouldn't r3=1/[tex]\sqrt{2}[/tex] and [tex]\theta[/tex]=-[tex]\pi[/tex]/36 ..but that's only 1 solution?
 
Physics news on Phys.org
well.
if [itex]z^3=\frac{1}{\sqrt{2}} e^{-\frac{i \pi}{12}}[/itex]
then [itex]z=(\frac{1}{\sqrt{2}})^{\frac{1}{3}} e^{-\frac{i \pi}{36}}[/itex]
 
but that is only one solution and it asks for 3
 
You can add any multiple of 2 pi i to theta.
 
ahh, of course. thank you!
 
EmmaK said:
ahh, of course. thank you!

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

Otherwise you're just describing the same number over and over!
 
Last edited:
haha, oh yea.
where do you get n2/pi i from?
 
EmmaK said:
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, [tex]r e^{i\theta}[/tex] so that you get the same value.

[tex]re^{i\theta}=re^{i(\theta+2\pi)}[/tex]

You can easily see this using Euler's identity since the sine and cosine both have a period of [tex]2\pi[/tex] 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 [tex]2\pi[/tex] once, twice, or three times to the original exponent's angle.
 
yes i should have mentioned that. sorry.
 
so the final answer is [tex]\stackrel{n2\pi}{3}[/tex]? i think i just misread your post as [tex]\stackrel{2n}{\pi}[/tex] or something :)
 
well, is [itex](\frac{2n \pi}{3})^3=\frac{1}{\sqrt{2}}e^{-\frac{1 \pi}{12}}[/itex]?

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