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Welcome to NRICH.

Hyperbolic functions, natural logs and complex numbers

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By Nikhil Shah (P3372) on Sunday, December 10, 2000 - 02:27 pm:

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)

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By Michael Doré (Md285) on Sunday, December 10, 2000 - 04:08 pm:

Hi,

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

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By Nikhil Shah (P3372) on Sunday, December 10, 2000 - 09:06 pm:

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!

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By Kerwin Hui (Kwkh2) on Sunday, December 10, 2000 - 11:06 pm:

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

Kerwin

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By Kerwin Hui (Kwkh2) on Sunday, December 10, 2000 - 11:37 pm:

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

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