My Math Forum Solve (e^z-1)^3 = 1

 Complex Analysis Complex Analysis Math Forum

 February 5th, 2017, 12:17 PM #1 Newbie   Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0 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
 February 5th, 2017, 12:51 PM #2 Senior Member     Joined: Sep 2015 From: Southern California, USA Posts: 1,604 Thanks: 817 $\left(e^{(z-1)}\right)^3=1$ $\left(e^{z-1}\right)^3 = e^{i 2\pi k},~k\in \mathbb{Z}$ $\left(e^{z-1}\right)=e^{i \frac{2\pi k}{3}},~k\in \mathbb{Z}$ $z-1 = i \dfrac{2\pi k}{3},~k\in \mathbb{Z}$ $z = 1+i \dfrac{2\pi k}{3},k \in \mathbb{Z}$ Thanks from topsquark
February 5th, 2017, 01:21 PM   #3
Newbie

Joined: Jan 2017

Posts: 21
Thanks: 0

Quote:
 Originally Posted by romsek $\left(e^{(z-1)}\right)^3=1$ $\left(e^{z-1}\right)^3 = e^{i 2\pi k},~k\in \mathbb{Z}$ $\left(e^{z-1}\right)=e^{i \frac{2\pi k}{3}},~k\in \mathbb{Z}$ $z-1 = i \dfrac{2\pi k}{3},~k\in \mathbb{Z}$ $z = 1+i \dfrac{2\pi k}{3},k \in \mathbb{Z}$

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

Last edited by skipjack; February 5th, 2017 at 07:22 PM.

February 5th, 2017, 02:00 PM   #4
Senior Member

Joined: Sep 2015
From: Southern California, USA

Posts: 1,604
Thanks: 817

Quote:
 Originally Posted by Lalaluye Oh no, sorry it's ((e^z)- 1) = 1
No cubed?

Last edited by skipjack; February 5th, 2017 at 07:23 PM.

February 5th, 2017, 02:01 PM   #5
Newbie

Joined: Jan 2017

Posts: 21
Thanks: 0

Quote:
 Originally Posted by romsek 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.

February 5th, 2017, 02:21 PM   #6
Senior Member

Joined: Sep 2015
From: Southern California, USA

Posts: 1,604
Thanks: 817

Quote:
 Originally Posted by Lalaluye 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.

 February 5th, 2017, 04:47 PM #7 Newbie   Joined: Jan 2017 From: Canada Posts: 21 Thanks: 0 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.
February 5th, 2017, 06:46 PM   #8
Senior Member

Joined: Sep 2015
From: Southern California, USA

Posts: 1,604
Thanks: 817

Quote:
 Originally Posted by Lalaluye 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

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

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

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

Last edited by romsek; February 5th, 2017 at 07:25 PM.

 February 5th, 2017, 07:21 PM #9 Global Moderator   Joined: Dec 2006 Posts: 18,155 Thanks: 1422 Haven't you omitted something? Thanks from romsek

 Tags ez13, solve

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post gen_shao Algebra 12 November 2nd, 2014 07:11 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top