Finding Arctan of a Complex NumberDate: 08/18/2008 at 14:19:29 From: BCox Subject: Arctan(x+iy)= Real() + Im() Can we separate the real and imaginary parts of arctan(x+iy) expression? I am not sure how to answer this using Asin/acos/atan for Complex Numbers http://mathforum.org/library/drmath/view/52235.html We have that Arctan(x+iy) = -i*Arctanh(ix-y) This is only useful if we already knew how to separate the real from the imaginary part of Arctanh(ix-y). so that if theta = arctan( x+iy / v+iw ) = arctan [ (x+iy)(v-iw) / (v^2+w^2) ] = arctan [ (xv+yw)/(v^2+w^2) + i (yv-wx)/(v^2+w^2) ] = ? (i/2)log{ (xv+yw)/(v^2+w^2) + i* [1+(yv-wx)/(v^2+w^2)] } -(i/2)log{ -(xv+yw)/(v^2+w^2) + i* [1-(yv-wx)/(v^2+w^2)] } Date: 08/23/2008 at 15:49:37 From: Doctor Vogler Subject: Re: Arctan(x+iy)= Real() + Im() Hi, Thanks for writing to Dr. Math. Using the method from the article you referenced, we would solve for arctan(x+iy) as follows. First let's write z = x + iy and do the first part in complex numbers, and we'll also write w = arctan(x + iy) so that tan(w) = z and we want to solve for w. So first we use sin(w) = (e^(iw) - e^(-iw))/(2i) and cos(w) = (e^(iw) + e^(-iw))/2 to get tan(w) = (e^(iw) - e^(-iw))/(i(e^(iw) + e^(-iw))) Letting p = e^(iw), so that 1/p = e^(-iw), the equation tan(w) = z becomes (p - 1/p)/(i(p + 1/p)) = z (p^2 - 1)/(p^2 + 1) = iz p^2 - 1 = iz(p^2 + 1) (1 - iz)p^2 = 1 + iz p^2 = (1 + iz)/(1 - iz) which is equivalent to e^(2iw) = (1 + iz)/(1 - iz). So the solution for w is w = 1/(2i) * ln ((1 + iz)/(1 - iz)). Now, to get the real and imaginary parts of this, you need the real and imaginary parts of the natural logarithm. You do this by thinking in terms of converting between polar and rectangular coordinates, since e^(x + iy) = e^x * cos y + i * e^x * sin y. Namely, the real part of ln(z) is ln |z|, the natural logarithm of the absolute value (a.k.a. norm or modulus) of z, or re(ln(z)) = ln |z| = ln sqrt(re(z)^2 + im(z)^2) = (1/2) ln (re(z)^2 + im(z)^2). The imaginary part is the angle at which z is found, which is either arctan( im(z)/re(z) ), plus any integer multiple of 2*pi, when re(z) > 0, or pi + arctan( im(z)/re(z) ), plus any integer multiple of 2*pi, when re(z) < 0; when re(z) = 0, it is pi/2 when im(z) > 0, or -pi/2 when im(z) < 0. You can alternately compute pi/2 - arctan( re(z)/im(z) ), when im(z) > 0, or -pi/2 - arctan( re(z)/im(z) ), when im(z) < 0, and then when im(z) = 0, you take 0 for re(z) > 0 and pi for re(z) < 0. Now, let's put all this together: First you have z = x + iy and you need to take the log of (1 + iz)/(1 - iz) = (1 + ix - y)/(1 - ix + y) = (1 - y + ix)(1 + y + ix)/((1 + y)^2 + x^2) = (1 - x^2 - y^2 + 2ix)/((1 + y)^2 + x^2). So the real part of the log is (1/2) ln ( (1 - x^2 - y^2)^2 + (2x)^2 ) - ln ( (1 + y)^2 + x^2 ). Since the sign of the imaginary part of the complex number whose log you are taking is the same as the sign of x, the imaginary part of the log is pi/2 - arctan ( (1 - x^2 - y^2)/(2x) ) when x > 0, or -pi/2 - arctan ( (1 - x^2 - y^2)/(2x) ) when x = 0, or 0 when x = 0 and 1 > x^2 + y^2, or pi when x = 0 and 1 < x^2 + y^2, plus any integer multiple of 2*pi. Finally, we multiply this log by 1/(2i), and we get im(arctan(x + iy)) = -(1/4) ln ((1 - x^2 - y^2)^2 + (2x)^2) + (1/2) ln ((1 + y)^2 + x^2) and re(arctan(x + iy)) = pi/4 - (1/2) arctan ( (1 - x^2 - y^2)/(2x) ) when x > 0, or re(arctan(x + iy)) = -pi/4 - (1/2) arctan ( (1 - x^2 - y^2)/(2x) ) when x < 0, or re(arctan(x + iy)) = 0 when x = 0 and -1 < y < 1, or re(arctan(x + iy)) = pi/2 when x = 0 and 1 < y^2, where each of these real parts can be increased by any integer multiple of pi (which makes perfect sense, since it is well-known that tan(x + n*pi) = tan(x)). Note that tan(z) can never equal i or -i, so arctan(i) and arctan(-i) are undefined. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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