# De Moivre's theorem

Hi guys, just a simple question.

I need to find all solutions of z^6 = 1.

I need to change that to |z|^6(cos 6θ + isin6θ) = 1 and the prof wrote → 1(1+0i). How did the prof do this? The rest of the question I understand. It's just that one step still stumps me. Thanks

jambaugh
Gold Member
I'm not clear on your question. Are you asking why 1 = 1+0i? or 1 = 1*1 = 1*(1+0i)?

Note: A more compact notation is: $r\cdot(\cos(\phi) + i\sin(\phi)) = r e^{i\phi}$. (This is the analytic extension of the exponential function to the complex plane, the real and imaginary components of $e^{i\phi}$ are the cosine and sine functions respectively.)

In this notation: $1 = 1\cdot e^{0 i} = 1\cdot e^{(0 + 2\pi k)i}$ for $k = 0, 1, 2, 3, \ldots$

Then all solutions can be obtained by taking the 6th root of all forms of 1:
$z = [ e^{0 + 2\pi k}]^{1/6} = e^{\frac{2 \pi}{6} k} = e^{\frac{\pi}{3} k} = cos(\frac{\pi}{3} k ) + i \sin(\frac{\pi}{3} k)$
iterate through values of k until you begin repeating.

For that question, here is the answer the prof wrote.
Find all the solutions of Z^6 = 1.

|z|^6(cos6θ + isin6θ)

|z|^6 = 1
|z| = 1
Therefore:
1(1 +0i) <-- don't get how the prof got to here.

cos6θ = 1 sin6θ = 0

6θ = 0+n(2π)
= n(π/3)

so 0, π/3, 2π/3, π, 4π/3, 5π/3

z1 = (cos0 + isin0) = 1
Blah Blah Blah..

jambaugh
Gold Member
For that question, here is the answer the prof wrote.
Find all the solutions of Z^6 = 1.

|z|^6(cos6θ + isin6θ)

|z|^6 = 1
|z| = 1
Therefore:
1(1 +0i) <-- don't get how the prof got to here.
Again I'm not clear on your question.
Are you saying you have difficulty understanding why $1 = 1+0i?$ or $1 = 1\cdot(1) = 1\cdot(1+0i)$?

If not let me fill in some gaps in the exposition.
-----------------------------------------------------
Find all solutions to $Z^6 = 1$.

DeMoivre's theorem states that for $Z = |Z|(\cos(\theta) + i\sin(\theta) )$, we have $Z^n = |Z|^n (\cos(n\theta) + i\sin(n\theta))$.

Apply this here by assuming $Z = |Z|(\cos(\theta) + i\sin(\theta) )$ and setting
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1$ which in polar form is $1 = 1(1 + 0i )$

First solve for the magnitude... (magnitude of l.h.s. must equal magnitude of r.h.s.):

Since $|Z|^6 = |1| = 1$ you have $|Z| = \sqrt{1}=1$.

Now apply the known magnitude of Z...

therefore $Z^6 = 1( \cos(6\theta) + i\sin(6\theta)) = 1(1+ 0i)$
and thus $\cos(6\theta) = 1, \sin(6\theta) = 0$...

the rest if solving for the thetas such that this is true.

Sorry for being unclear, but I think it's the step in your explanation that turning the De Moivre theorem to polar form. How do you do that?

Again I'm not clear on your question.
Are you saying you have difficulty understanding why $1 = 1+0i?$ or $1 = 1\cdot(1) = 1\cdot(1+0i)$?

If not let me fill in some gaps in the exposition.
-----------------------------------------------------
Find all solutions to $Z^6 = 1$.

DeMoivre's theorem states that for $Z = |Z|(\cos(\theta) + i\sin(\theta) )$, we have $Z^n = |Z|^n (\cos(n\theta) + i\sin(n\theta))$.

Apply this here by assuming $Z = |Z|(\cos(\theta) + i\sin(\theta) )$ and setting
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1$ which in polar form is $1 = 1(1 + 0i )$ <- This part is unclear to me

First solve for the magnitude... (magnitude of l.h.s. must equal magnitude of r.h.s.):

Since $|Z|^6 = |1| = 1$ you have $|Z| = \sqrt{1}=1$.

Now apply the known magnitude of Z...

therefore $Z^6 = 1( \cos(6\theta) + i\sin(6\theta)) = 1(1+ 0i)$
and thus $\cos(6\theta) = 1, \sin(6\theta) = 0$...

the rest if solving for the thetas such that this is true.

jambaugh
Gold Member
Sorry for being unclear, but I think it's the step in your explanation that turning the De Moivre theorem to polar form. How do you do that?

De Moivre's theorem is already in polar form.

Are you saying you don't understand that 1 = x + iy with x = 1 and y = 0?

Can't you plot complex numbers in the complex plane? Plot 1 in the complex plane. Where is it? At what angle and how far from the origin?

EDIT:
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1$
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 + 0$
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1+ 0 i$
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) =1( 1+ 0 i )$

I don't understand why Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 equals to 1( 1+0i)

De Moivre's theorem is already in polar form.

Are you saying you don't understand that 1 = x + iy with x = 1 and y = 0?

Can't you plot complex numbers in the complex plane? Plot 1 in the complex plane. Where is it? At what angle and how far from the origin?

EDIT:
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1$
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 + 0$
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1+ 0 i$
$Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) =1( 1+ 0 i )$

jambaugh
Gold Member
I don't understand why Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 equals to 1( 1+0i)

Which = in this equation don't you understand?

The first is just setting n=6 in De Moivre's theorem. Do you have De Moivre's theorem in front of you? Pull out your textbook and look at it.

The 2nd "=" is the original equation you're trying to solve.

The 3rd "=" is OBVIOUS. Each step I showed in the last post is an identity substitution. 1=1+ 0 because 0 is the additive identity.
0i = 0 by a very basic algebraic lemma.
(1+0i) = 1(1+0i) because 1 is the multiplicative identity.

I'm going to bed now. If you can be more explicit in exactly what it is you do not understand, rather than just pointing at a step, I'll leave it to others to enlighten you.

EDIT: Sorry about being so terse. I was sleepy and cranky. I've slept now and am neither. So please ignore/forgive my rudeness. Is there anything I can do to clarify this further?

Last edited:
would u mind can u answer this for me... please EVALUATE: e^e^m(pi+i pi/6)?

jambaugh
$$z=e^{m\pi (1+i)/6} = e^{m\pi/6}[\cos(m\pi/6) + i \sin(m\pi /6)] = x+iy$$
$$e^z = e^{x+iy} =e^x\cdot e^{iy}= e^x(\cos(y)+i\sin(y))$$