Whats wrong here?

  • Thread starter fargoth
  • Start date
  • #1
315
6
[tex]sin(z)=2[/tex]
[tex]e^{iz}-e^{-iz} = 4i[/tex]
[tex]e^{2iz}-4ie^{iz} = 1[/tex]
[tex]iz \ln (e^{iz}-4i) = 0[/tex]
[tex] z=0[/tex]

when solving it by
[tex]w = e^{iz}[/tex]
[tex]w^2-4wi-1 = 0[/tex]
i get one more solution, why is the first way not as good as the second way?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956
ln(ab)= ln(a)+ ln(b), not ln(a)*ln(b).

If [tex]e^{2iz}-4ie^{iz} = 1[/tex]
then [tex]e^{iz}(e^{iz}- 4i)= 1[/tex]
so [tex]iz+ ln(e^{iz}- 4i)= 0[/tex]
NOT [tex]iz \ln (e^{iz}-4i) = 0[/tex]
 
  • #3
315
6
hehe, right, that was a dumb mistake :biggrin:
thank you for pointing it out.
 

Related Threads on Whats wrong here?

Replies
6
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
673
Replies
1
Views
943
Replies
1
Views
823
  • Last Post
Replies
4
Views
1K
Replies
2
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
1K
Top