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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

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- #1

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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

- #2

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

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

Then all solutions can be obtained by taking the 6th root of all forms of 1:

[itex] 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)[/itex]

iterate through values of k until you begin repeating.

- #3

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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..

- #4

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Again I'm not clear on your question.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.

If not let me fill in some gaps in the exposition.

-----------------------------------------------------

Find all solutions to [itex] Z^6 = 1[/itex].

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

Apply this here by assuming [itex]Z = |Z|(\cos(\theta) + i\sin(\theta) )[/itex] and setting

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1[/itex] which in polar form is [itex]1 = 1(1 + 0i )[/itex]

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

Since [itex] |Z|^6 = |1| = 1[/itex] you have [itex] |Z| = \sqrt[6]{1}=1[/itex].

Now apply the known magnitude of Z...

and thus [itex] \cos(6\theta) = 1, \sin(6\theta) = 0[/itex]...

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

- #5

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Again I'm not clear on your question.

Are you saying you have difficulty understanding why [itex]1 = 1+0i?[/itex] or [itex]1 = 1\cdot(1) = 1\cdot(1+0i)[/itex]?

If not let me fill in some gaps in the exposition.

-----------------------------------------------------

Find all solutions to [itex] Z^6 = 1[/itex].

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

Apply this here by assuming [itex]Z = |Z|(\cos(\theta) + i\sin(\theta) )[/itex] and setting

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1[/itex] which in polar form is [itex]1 = 1(1 + 0i )[/itex]<- This part is unclear to me

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

Since [itex] |Z|^6 = |1| = 1[/itex] you have [itex] |Z| = \sqrt[6]{1}=1[/itex].

Now apply the known magnitude of Z...

therefore [itex]Z^6 = 1( \cos(6\theta) + i\sin(6\theta)) = 1(1+ 0i)[/itex]

and thus [itex] \cos(6\theta) = 1, \sin(6\theta) = 0[/itex]...

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

- #6

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De Moivre's theorem is already in polar form.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?

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:

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1[/itex]

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 + 0[/itex]

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1+ 0 i[/itex]

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) =1( 1+ 0 i )[/itex]

- #7

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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:

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1[/itex]

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 + 0[/itex]

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1+ 0 i[/itex]

[itex]Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) =1( 1+ 0 i )[/itex]

- #8

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Which = in this equation don't you understand?I don't understand why Z^6 = |Z|^6(\cos(6\theta) + i\sin(6\theta) ) = 1 equals to 1( 1+0i)

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

0i = 0 by a very basic algebraic lemma.

(1+0i) = 1(1+0i) because 1 is the

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:

- #9

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

- #10

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Tell me the value of m.would u mind can u answer this for me... please EVALUATE: e^e^m(pi+i pi/6)?

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

[tex]e^z = e^{x+iy} =e^x\cdot e^{iy}= e^x(\cos(y)+i\sin(y))[/tex]

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