# Homework Help: Complex Numbers homework

1. Aug 26, 2007

### chaoseverlasting

1. The problem statement, all variables and given/known data

Express $$cos^{-1}(x+iy)$$ in the form $$A+iB$$).

3. The attempt at a solution

$$x+iy=cos(a+ib)$$
$$x-iy=cos(a-ib)$$
$$2x=2cos(a)cosh(b)$$
$$x=cosa coshb$$
Similarly,
$$y=-sina sinhb$$

Using these values, I got $$x^2+y^2=cos^2a +sinh^2b$$, but I dont know where to go from here.

Alternatively,
$$a+ib=cos^{-1}(x+iy)$$
$$a-ib=cos^{-1}(x-iy)$$
$$2a=cos^{-1}(x^2+y^2 -\sqrt{1-(x+iy)^2}\sqrt{1-(x-iy)^2})$$
and similarly,
$$2b=cos{-1}(x^2+y^2+\sqrt{1-(x+iy)^2}\sqrt{1-(x-iy)^2}$$,

but after expanding, these expressions are too complex. Is this the final expression though? I dont have the answer, so I have nothing to compare this to.

2. Aug 26, 2007

### Hurkyl

Staff Emeritus
That's not right.

3. Aug 27, 2007

### chaoseverlasting

Why though? Here's what I did:

$$x+iy=cos(a+ib)$$
$$x+iy=cosa cos(ib) -sina sin(ib)$$
$$x+iy=cosa coshb -i sina sinhb$$ (as cos(ib)=cosh b and sin(ib)=i sinhb)
Equating the real and imaginary parts,
$$x=cos a cosh b$$
$$y=-sina sinhb$$

Squaring and adding,

$$x^2+y^2=cos^2acosh^2b +sin^2a sinh^2b$$
As $$cosh^2a=1+sinh^2b$$,

$$x^2+y^2=cos^2a +sinh^2b(cos^2a+sin^2a)$$
Hence,
$$x^2+y^2=cos^2a +sinh^2b$$.

What did I do wrong? Where do I go from here?

4. Aug 30, 2007

### chaoseverlasting

Anyone got any hints/ideas on this one?

5. Aug 30, 2007

### learningphysics

I'd do it like this:

if z = A + iB

then use the definition of cos...

$$\frac{e^{iz} + e^{-iz}}{2} = x + iy$$

Then I'd let $$h = e^{iz}$$

So you have

$$\frac{h+1/h}{2} = x + iy$$

So first solve for h... then get z out of that... then get A and B from that...

I didn't actually work through this, so I don't know if it will work... just an idea...

6. Aug 30, 2007

### HallsofIvy

What reason do you have for thinking that if x+ iy= cos(a+ bi) then
x- iy= cos(a-bi)?

I think I would try to use
$$cos(z)= \frac{e^z+ e^{-z}}{2}$$
Of course, if z= x+ iy then
$$e^{x+iy}= e^x(cos(y)+ i sin(y))$$
and
$$e^{-(x-iy)}= e^{-x}x(cos(y)- i sin(y))$$
so that is
$$cos(x+iy)= \frac{e^x(cos(y)+ i sin(y))+e^{-x}(cos(y)- i sin(y))}{2}$$
Now separate the real and imaginary parts of that.

7. Aug 30, 2007

### learningphysics

But he needs cos^-1(x+iy)...

Is this derivation to show that x+ iy= cos(a+ bi) implies x- iy= cos(a-bi) ?

8. Aug 30, 2007

### phoenixthoth

That's a typo, right?

There could be two ways to correct it...

$$cosh(z)= \frac{e^z+ e^{-z}}{2}$$

OR

$$cos(z)= \frac{e^{i z}+ e^{-i z}}{2}$$

9. Aug 30, 2007

### D H

Staff Emeritus
It pops right out of the identity,
$$\cos(a+ib) = \cos a \cosh b - i \sin a \sinh b$$

10. Aug 31, 2007

### Gib Z

Letting $$\cos z = y = \frac{e^{iz} + e^{-iz}}{2}$$, replace all x's with y's and y's with x's, which is what you do to find inverse relations. Then solve for y using logs.

11. Aug 31, 2007

### chaoseverlasting

Actually, I solved it. Here's what I did:

$$x+iy=cos(a+ib)$$
$$x-iy=cos(a-ib)$$
$$cos(2a)=cos(a+ib+a-ib)=x^2+y^2-\sqrt{(1-(x+iy)^2)(1-(x-iy)^2)}$$ (using cos(a+b))
$$cos(2a)=x^2+y^2+(x^2+y^2-1)$$ or$$cos(2a)=x^2+y^2-(x^2+y^2-1)$$

This gives you $$A=(2n+1)\frac{\pi}{4} or A=\frac{1}{2}cos^{-1}(2x^2+2y^2-1)$$

Similarly,
$$B=\frac{1}{2}cosh^{-1}(2x^2+2y^2-1) or iB=(2n+1)\frac{\pi}{4}$$. But which is it? How do I eliminate one solution set?

Last edited: Aug 31, 2007
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