hold on, david. the formula I posted previously from wolfram is ArcTan[x,y] with x or y complex: its the same of arctan2(x,y). arctan is another function (even though arctan2(y,x) should be "a better" arctan(y/x)).
the correct formula for y = arctan(x), with any x (real or complex), should be (if I still can play with sin and cos...): y = arctan(x) = 1/(2j) * log((1j-x)/(1j+x)) [ you can get it doing: y = arctan(x) --> x = tan(y) = sin(x)/cos(x) = -1j * (exp(1j*y)-exp(-1j*y))/(exp(1j*y)+exp(-1j*y); then let z = exp(1j*y) and solve in z.] I've tested the formula and it seems ok for different inputs (I've checked that tan(arctan(x)) == x): --------------------------- octave:56> x = 1; tan(1/2/1j*log((1j-x)/(1j+x))) ans = 1.0000 octave:57> x = 1j; tan(1/2/1j*log((1j-x)/(1j+x))) ans = -0 + 1i octave:58> x = 2j; tan(1/2/1j*log((1j-x)/(1j+x))) ans = 1.8369e-16 + 2.0000e+00i octave:59> x = 1+2j; tan(1/2/1j*log((1j-x)/(1j+x))) ans = 1.0000 + 2.0000i --------------------------- hth, L. On 4/30/07, David Goldsmith <[EMAIL PROTECTED]> wrote:
> (hint what is arctan(0+1j)?) > Well, at the risk of embarrassing myself, using arctan(x+iy) = I get: arctan(0+1i) = -i*log((0+i*1)/sqrt(0^2 + 1^2)) = -i*log(i/1) = -i*log(i) = -i*log(exp(i*pi/2)) = -i*i*pi/2 = pi/2... Is there some reason I'm forgetting (e.g., a branch cut convention or something) why this is wrong? DG _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
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