Complex Analysis: Show RHS=LHS for Real r?

[jjou]

[SOLVED] Complex Analysis

Show that $$\mbox{Re}\left(\frac{Re^{i\theta}+r}{Re^{i\theta}-r}\right)=\frac{R^2-r^2}{R^2-2Rr\cos\theta+r^2}$$ 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 $$Re^{i\theta}=R\cos\theta+iR\sin\theta$$ and $$r=a+ib$$, I got

$$\frac{R^2-a^2-b^2}{R^2-2R(a\cos\theta+b\sin\theta)+a^2+b^2}$$

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, $$r^2$$ in the numerator and $$r^2$$ is, in general, not real for r not real.

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

Thanks. :)

 

[jhicks]

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 $\theta$ would be replaced with something like $\theta + \angle r$ (as you found using rectangular form).

 

[jjou]

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.

 

[jhicks]

You showed it was false. Just by inspection of the numerator, $r^2 \neq a^2 + b^2 = |r|^2$ for complex r

 

[jjou]

Thanks. :)