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February 5th, 2017, 12:17 PM   #1
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Solve (e^z-1)^3 = 1

How do you solve for

(e^z-1)^3 = 1

I expanded this out and got this far:

e^3z - 3e^2z + 3e^z -2 = 0
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February 5th, 2017, 12:51 PM   #2
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$\left(e^{(z-1)}\right)^3=1$

$\left(e^{z-1}\right)^3 = e^{i 2\pi k},~k\in \mathbb{Z}$

$\left(e^{z-1}\right)=e^{i \frac{2\pi k}{3}},~k\in \mathbb{Z}$

$z-1 = i \dfrac{2\pi k}{3},~k\in \mathbb{Z}$

$z = 1+i \dfrac{2\pi k}{3},k \in \mathbb{Z}$
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February 5th, 2017, 01:21 PM   #3
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Quote:
Originally Posted by romsek View Post
$\left(e^{(z-1)}\right)^3=1$

$\left(e^{z-1}\right)^3 = e^{i 2\pi k},~k\in \mathbb{Z}$

$\left(e^{z-1}\right)=e^{i \frac{2\pi k}{3}},~k\in \mathbb{Z}$

$z-1 = i \dfrac{2\pi k}{3},~k\in \mathbb{Z}$

$z = 1+i \dfrac{2\pi k}{3},k \in \mathbb{Z}$

Oh no, sorry it's ((e^z)- 1) = 1

Last edited by skipjack; February 5th, 2017 at 07:22 PM.
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February 5th, 2017, 02:00 PM   #4
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Quote:
Originally Posted by Lalaluye View Post
Oh no, sorry it's ((e^z)- 1) = 1
No cubed?

Last edited by skipjack; February 5th, 2017 at 07:23 PM.
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February 5th, 2017, 02:01 PM   #5
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Quote:
Originally Posted by romsek View Post
No cubed?
Ah! Another typo

it's ((e^z) - 1)^3 = 1

This is the correct one.

Last edited by skipjack; February 5th, 2017 at 07:23 PM.
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February 5th, 2017, 02:21 PM   #6
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Quote:
Originally Posted by Lalaluye View Post
Ah! Another typo

it's ((e^z) - 1)^3 = 1

This is the correct one.
You can't just apply what I did to this? It's pretty straightforward.

Last edited by skipjack; February 5th, 2017 at 07:24 PM.
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February 5th, 2017, 04:47 PM   #7
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This is what I got.

I'm not sure whether it is correct.

$\displaystyle ((e^z) - 1)= e^i(2pik/3)$
$\displaystyle e^z - e^i(2pi) = e^i(2pik/3)$
$\displaystyle e^z = e^i(2pik/3) + e^i(2pi)$
$\displaystyle z = i(2pik/3) + i(2pi)$
$\displaystyle z = i(2pik/3 + 2 pi)$


Is this right?

Last edited by skipjack; February 5th, 2017 at 07:25 PM.
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February 5th, 2017, 06:46 PM   #8
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Quote:
Originally Posted by Lalaluye View Post
Ah! Another typo

its ((e^z)- 1)^3 = 1

This is the correct one
$\left(e^z - 1\right)^3 = 1$

$e^z - 1 = e^{\frac{i 2 \pi k}{3}}$

$e^z - 1 = \cos\left(\frac{ 2 \pi k}{3}\right) + i \sin\left(\frac{ 2 \pi k}{3}\right)$

$e^z = 1+\cos\left(\frac{ 2 \pi k}{3}\right) + i \sin\left(\frac{ 2 \pi k}{3}\right)$

Noting that there are only 3 values of $k$ that produce unique values of the right hand side we can go ahead and expand it.

$e^z = 2$

$e^z = \dfrac 1 2+i \dfrac{\sqrt{3}}{2}$

$e^z = \dfrac 1 2-i \dfrac{\sqrt{3}}{2}$

we can see by inspection that

$z = \ln(2) +i 2\pi k,~k \in \mathbb{Z}$

$z = i\left(\dfrac{\pi}{3}+2\pi k\right), ~k \in \mathbb{Z}$

$z = -i\left(\dfrac{\pi}{3}+2\pi k\right), ~k \in \mathbb{Z}$
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Last edited by romsek; February 5th, 2017 at 07:25 PM.
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February 5th, 2017, 07:21 PM   #9
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Haven't you omitted something?
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