Hi, I've tried several approaches to this question, but just kept on ending up with a ridiculous blurb of algebra! Could someone please help me?!

If u+iv = ln (x+iy+a / x+iy-a), prove that:

a) x² + y² -2axcothu + a² = 0

b) x = (asinhu) / (coshu -cosv)

c) |x+iy|² = a² (coshu + cosv) / (coshu -cosv)

Hi,

I'm not quite sure I understand. Are x,y,u,v real? What about a - is that real or complex?

Well, all the letters used are real, since anything that is imaginary has already got an i in front of it. this is the question just as it is written out in the book, so I assume that this is the case - it make's sense!

The substitution z=x+iy and w=u+iv should simplify a lot of the algebra. Obviously there is some manipulation....

Kerwin

OK, I think I'll write a brief summary of what to do. Let z=x+iy, w=u+iv, we have w=ln [(z+a)/(z-a)]

Rearrange to get

z=a(e^w +1)/(e^w -1)

Recall: |z|² =x² +y² =z z^* , where z^* is the complex conjugate of z.

So after a bit of algebra (about 4 lines), we have

|z|² =a² [(e^u +e^-u +e^iv +e^-iv )/(e^u +e^-u -e^iv -e^-iv )]

which is part c), as e^iv +e^-iv is 2cos v, etc.

Also, from z=a(e^w +1)/(e^w -1), we have

z=a{[e^u -e^-u -(e^iv -e^-iv )]/[e^u +e^-u -(e^iv +e^-iv )]}

which yields b) when you equate the real parts.

Now from c) and b) we get a).

Kerwin