[Numpy-discussion] arctan2 with complex args
David Goldsmith
David.L.Goldsmith@noaa....
Mon Apr 30 10:26:49 CDT 2007
lorenzo bolla wrote:
> 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))
After a little algebra, they're the same formula (as they must be; hint:
"rationalize" the denominator inside the log and bring the 1/2 inside
the log, making it a square root, and finally, rewrite 1/i as -i); I
have to run my kid to school right now, but if you want me to flesh this
out, reply and I'll type it out and send it when I get back..
DG
>
> [ 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* <David.L.Goldsmith@noaa.gov
> <mailto:David.L.Goldsmith@noaa.gov>> 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
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