P.S. Out of curiosity, I just plotted the surface z = (x^2 - y^2) / (x^2 + y^2) which is of course just z = cos(2 theta), and it gives a cute graph with a vertical interval as singular points at (x,y) = (0,0). This is well-known to some people, but it wasn't to me. You get essentially the identical graph for z = 2xy / (x^2 + y^2) = sin(2 theta). --Dan On 2013-04-26, at 1:06 PM, Dan Asimov wrote:
The same thing also comes out from
|z| = |conj(z)| =>
1^2 = |K + iL|^2 / |K - iL|^2 = |(K + iL) / (K - iL)|^2
= |(K + iL)^2 / (K^2 + L^2)|^2
= |(K^2 - L^2) + i2KL|^2 / (K^2 + L^2)^2
--Dan
On 2013-04-26, at 1:04 AM, Bill Gosper wrote:
In[570]:= (1+2*I)/(2+1*I) Out[570]= 4/5+(3 I)/5
In[571]:= (1+3*I)/(3+1*I) Out[571]= 3/5+(4 I)/5
In[572]:= (2+3*I)/(3+2*I) Out[572]= 12/13+(5 I)/13
In[573]:= (1+4*I)/(4+1*I) Out[573]= 8/17+(15 I)/17
In[586]:= Simplify/@ComplexExpand[(a+I*b)/(b+I*a)] Out[586]= (2 a b)/(a^2+b^2)-(I (a^2-b^2))/(a^2+b^2)
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun