Finding Imaginary Numbers with Cosine of 3

[Zurtex]

Question:

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

My workings so far:

cos(x)cos(iy) - sin(x)sin(iy) = 3
Therefore:
cos(x)cosh(y) - i(sin(x)sinh(y)) = 3

cos(x)cosh(y) = 3
sin(x)sinh(y) = 0

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

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.

 

[NateTG]

I'm not so sure the identities that you're using a valid, but you have:
<br /> \sin(x)\sin(y)=0 \rightarrow \sin(x)=0<br />
when \sin(y)=0 is also a possibility.

 

[Zurtex]

Originally posted by NateTG
I'm not so sure the identities that you're using a valid, but you have:
<br /> \sin(x)\sin(y)=0 \rightarrow \sin(x)=0<br />
when \sin(y)=0 is also a possibility.

I think you've misread my post.

\sin(x)\sinh(y)=0
not
\sin(x)\sin(y)=0

 

[NateTG]

Oh, ok.
There's also x=2\pi n for varius n right?

 

[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 cosh(ln(3 + 8^{\frac{1}{2}}))= 3 but it is also true that
cosh(-ln(3 + 8^{\frac{1}{2}}))= 3.

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

-ln(3 + 8^{\frac{1}{2}})i

 

[Zurtex]

Originally posted by NateTG
Oh, ok.
There's also x=2\pi n for varius n 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:

cos(iz) = cosh(z)
sin(iz) = i.sinh(z)
cosh(iz) = cos(z)
sinh(iz) = i.sin(z)

 

[Zurtex]

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 cosh(ln(3 + 8^{\frac{1}{2}}))= 3 but it is also true that
cosh(-ln(3 + 8^{\frac{1}{2}}))= 3.

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

-ln(3 + 8^{\frac{1}{2}})i

Great thanks

 

[NateTG]

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 z is imaginary, you have:
z=0+iy
so
3=\cos(z)=\cos(0+iy)=\cos(0)\cos(iy)+\sin(0)\sin(iy)=
\cos(iy)=\cosh(y)
so
3=\cosh(y)
so
3=\frac{e^y+e^{-y}}{2}
so let y&#039;=e^y
then
3=\frac{1}{2}(y+\frac{1}{y})
which solves to:
y&#039; \in \{3 + \sqrt{8}, 3 - \sqrt {8}\}
so
y \in \{\ln(3 + \sqrt{8}), \ln (3 - \sqrt{8})\}
which are the two solutions that you got since
-\ln(3 +\sqrt{8})=\ln(\frac{1}{3+\sqrt{8}})=\ln (3-\sqrt{8})