# Complex Roots

1. Feb 25, 2004

### 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.

2. Feb 25, 2004

### NateTG

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

3. Feb 25, 2004

### Zurtex

I think you've misread my post.

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

4. Feb 25, 2004

### NateTG

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

5. Feb 25, 2004

### HallsofIvy

Staff Emeritus
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

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

Last edited: Feb 25, 2004
6. Feb 25, 2004

### Zurtex

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:

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

7. Feb 25, 2004

### Zurtex

Great thanks

8. Feb 25, 2004

### 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 $$y'=e^y$$
then
$$3=\frac{1}{2}(y+\frac{1}{y})$$
which solves to:
$$y' \in \{3 + \sqrt{8}, 3 - \sqrt {8}\}$$
so
$$y \in \{\ln(3 + \sqrt{8}), \ln (3 - \sqrt{8})\}$$
which are the two solutions that you got since
$$-\ln(3 +\sqrt{8})=\ln(\frac{1}{3+\sqrt{8}})=\ln (3-\sqrt{8})$$

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