- #1

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. :)