Complex Roots

  • Thread starter Zurtex
  • Start date
  • #1
Zurtex
Science Advisor
Homework Helper
1,120
1
Question:

Use the formula
[tex]cos(x + iy) = cos(x)cos(iy) - sin(x)sin(iy)[/tex]
to find two imaginary numbers whose cosine is 3

My workings so far:

[tex]cos(x)cos(iy) - sin(x)sin(iy) = 3 [/tex]
Therefore:
[tex]cos(x)cosh(y) - i(sin(x)sinh(y)) = 3[/tex]

[tex]cos(x)cosh(y) = 3[/tex]
[tex]sin(x)sinh(y) = 0[/tex]

[tex]sin(x) = 0[/tex]
[tex]x = 0[/tex]
[tex]cos(x) = 1[/tex]
[tex]cosh(y) = 3[/tex]
[tex]y = ln(3 + 8^{\frac{1}{2}})[/tex]

However I can't seem to work out the other root. I know with polynomials it would would be its complex conjugate but I didn't want to assume so for other functions.
 

Answers and Replies

  • #2
NateTG
Science Advisor
Homework Helper
2,450
5
I'm not so sure the identities that you're using a valid, but you have:
[tex]
\sin(x)\sin(y)=0 \rightarrow \sin(x)=0
[/tex]
when [tex]\sin(y)=0[/tex] is also a possibility.
 
  • #3
Zurtex
Science Advisor
Homework Helper
1,120
1
Originally posted by NateTG
I'm not so sure the identities that you're using a valid, but you have:
[tex]
\sin(x)\sin(y)=0 \rightarrow \sin(x)=0
[/tex]
when [tex]\sin(y)=0[/tex] is also a possibility.
I think you've misread my post.

[tex]\sin(x)\sinh(y)=0[/tex]
not
[tex]\sin(x)\sin(y)=0[/tex]
 
  • #4
NateTG
Science Advisor
Homework Helper
2,450
5
Oh, ok.
There's also [tex]x=2\pi n[/tex] for varius [tex]n[/tex] right?
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
956
NateTG: other values for x, first, would not give "imaginary" answers. Since the problem said "imaginary", we could immediately let x= 0 and reduce to cosh(y)= 3. Secondly, other, non-zero values for x require that cosh(y) be less than 1 and that is not possible.


But cosh is an even function! Yes, it is true that [tex]cosh(ln(3 + 8^{\frac{1}{2}}))= 3[/tex] but it is also true that
[tex]cosh(-ln(3 + 8^{\frac{1}{2}}))[/tex]= 3.

The two imaginary numbers whose cosine is 3 are:
[tex]ln(3 + 8^{\frac{1}{2}})i[/tex] as you got and

[tex]-ln(3 + 8^{\frac{1}{2}})i[/tex]
 
Last edited by a moderator:
  • #6
Zurtex
Science Advisor
Homework Helper
1,120
1
Originally posted by NateTG
Oh, ok.
There's also [tex]x=2\pi n[/tex] for varius [tex]n[/tex] right?
I don't know, isn't that something to do with the roots of unity (something we are about to do). I've only just started this stopic and all we have taught so far is that:

[tex]cos(iz) = cosh(z)[/tex]
[tex]sin(iz) = i.sinh(z)[/tex]
[tex]cosh(iz) = cos(z)[/tex]
[tex]sinh(iz) = i.sin(z)[/tex]
 
  • #7
Zurtex
Science Advisor
Homework Helper
1,120
1
Originally posted by HallsofIvy
NateTG: other values for x, first, would not give "imaginary" answers. Since the problem said "imaginary", we could immediately let x= 0 and reduce to cosh(y)= 3. Secondly, other, non-zero values for x require that cosh(y) be less than 1 and that is not possible.


But cosh is an even function! Yes, it is true that [tex]cosh(ln(3 + 8^{\frac{1}{2}}))= 3[/tex] but it is also true that
[tex]cosh(-ln(3 + 8^{\frac{1}{2}}))[/tex]= 3.

The two imaginary numbers whose cosine is 3 are:
[tex]ln(3 + 8^{\frac{1}{2}})i[/tex] as you got and

[tex]-ln(3 + 8^{\frac{1}{2}})i[/tex]
Great thanks :smile: :smile: :smile:
 
  • #8
NateTG
Science Advisor
Homework Helper
2,450
5
Duh... Obviously I should not be posting before donuts and coffee.

As an aside here's a different way that both answers show up:

Since [tex]z[/tex] is imaginary, you have:
[tex]z=0+iy[/tex]
so
[tex]3=\cos(z)=\cos(0+iy)=\cos(0)\cos(iy)+\sin(0)\sin(iy)=[/tex]
[tex]\cos(iy)=\cosh(y)[/tex]
so
[tex]3=\cosh(y)[/tex]
so
[tex]3=\frac{e^y+e^{-y}}{2}[/tex]
so let [tex]y'=e^y[/tex]
then
[tex]3=\frac{1}{2}(y+\frac{1}{y})[/tex]
which solves to:
[tex]y' \in \{3 + \sqrt{8}, 3 - \sqrt {8}\}[/tex]
so
[tex]y \in \{\ln(3 + \sqrt{8}), \ln (3 - \sqrt{8})\}[/tex]
which are the two solutions that you got since
[tex] -\ln(3 +\sqrt{8})=\ln(\frac{1}{3+\sqrt{8}})=\ln (3-\sqrt{8})[/tex]
 

Related Threads on Complex Roots

  • Last Post
Replies
1
Views
1K
Replies
3
Views
8K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
13
Views
7K
Replies
14
Views
24K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
0
Views
2K
Top