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February 5th, 2017, 12:17 PM  #1 
Newbie Joined: Jan 2017 From: Canada Posts: 20 Thanks: 0  Solve (e^z1)^3 = 1
How do you solve for (e^z1)^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: CA Posts: 1,207 Thanks: 614 
$\left(e^{(z1)}\right)^3=1$ $\left(e^{z1}\right)^3 = e^{i 2\pi k},~k\in \mathbb{Z}$ $\left(e^{z1}\right)=e^{i \frac{2\pi k}{3}},~k\in \mathbb{Z}$ $z1 = i \dfrac{2\pi k}{3},~k\in \mathbb{Z}$ $z = 1+i \dfrac{2\pi k}{3},k \in \mathbb{Z}$ 
February 5th, 2017, 01:21 PM  #3  
Newbie Joined: Jan 2017 From: Canada Posts: 20 Thanks: 0  Quote:
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: CA Posts: 1,207 Thanks: 614  Last edited by skipjack; February 5th, 2017 at 07:23 PM. 
February 5th, 2017, 02:01 PM  #5 
Newbie Joined: Jan 2017 From: Canada Posts: 20 Thanks: 0  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: CA Posts: 1,207 Thanks: 614  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: 20 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: CA Posts: 1,207 Thanks: 614  $\left(e^z  1\right)^3 = 1$ $e^z  1 = e^{\frac{i 2 \pi k}{3}}$ $e^z  1 = \cos\left(\frac{ 2 \pi k}{3}\right) + i \sin\left(\frac{ 2 \pi k}{3}\right)$ $e^z = 1+\cos\left(\frac{ 2 \pi k}{3}\right) + i \sin\left(\frac{ 2 \pi k}{3}\right)$ 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. $e^z = 2$ $e^z = \dfrac 1 2+i \dfrac{\sqrt{3}}{2}$ $e^z = \dfrac 1 2i \dfrac{\sqrt{3}}{2}$ 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: 16,947 Thanks: 1255 
Haven't you omitted something?


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