How do I solve for cos^{-1}(x+iy) in the form A+iB?

[chaoseverlasting]

Homework Statement

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

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 don't 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 don't have the answer, so I have nothing to compare this to.

 

[Hurkyl]

x=\cos a \cosh b
y=-\sin a \sinh b

Using these values, I got x^2+y^2=\cos^2 a +\sinh^2 b, but I don't know where to go from here.

That's not right.

 

[chaoseverlasting]

That's not right.

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?

 

[chaoseverlasting]

Anyone got any hints/ideas on this one?

 

[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...

 

[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.

 

[learningphysics]

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.

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

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

 

[phoenixthoth]

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

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}

 

[D H]

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

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

 

[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.

 

[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) orcos(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?