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Complex Analysis

  • Thread starter jjou
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[SOLVED] Complex Analysis

Show that [tex]\mbox{Re}\left(\frac{Re^{i\theta}+r}{Re^{i\theta}-r}\right)=\frac{R^2-r^2}{R^2-2Rr\cos\theta+r^2}[/tex] where R is the radius of a disc.

I was able to show this for all real values of r. However, the problem doesn't specify whether r is real or complex. After expanding the left side using [tex]Re^{i\theta}=R\cos\theta+iR\sin\theta[/tex] and [tex]r=a+ib[/tex], I got

[tex]\frac{R^2-a^2-b^2}{R^2-2R(a\cos\theta+b\sin\theta)+a^2+b^2}[/tex]

which, as far as I know, is not equal to the right hand side of the above equation.

Intuitively, I also believe that this statement is not correct for non-real values of r since the LHS is a real value and the RHS has, at least, [tex]r^2[/tex] in the numerator and [tex]r^2[/tex] is, in general, not real for r not real.

Am I right in assuming this only holds for real-valued r?

Thanks. :)
 

Answers and Replies

334
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I can't see why r would denote a complex number considering the universal notation is z. Obviously the r's would be replaced by |r|'s in such a case and [itex]\theta[/itex] would be replaced with something like [itex]\theta + \angle r[/itex] (as you found using rectangular form).
 
Last edited:
64
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Right, but the question didn't specify, so I was worried...
Is that not necessary? Is it okay to assume that r always denotes a real number?

In any case, am I correct in saying that the statement is false for non-real values of r?

Thanks.
 
334
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You showed it was false. Just by inspection of the numerator, [itex]r^2 \neq a^2 + b^2 = |r|^2[/itex] for complex r
 
64
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Thanks. :)
 

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