Plotting Complex NumbersDate: 08/19/97 at 16:57:05 From: Cat Fyle Subject: Re: Modern Algebra I am still having some problems. I cannot figure out (1-i)^2i = 2^ie^1.570796 I think I probably will need to be walked through it. I have never had complex numbers. Thank you for all the help. Date: 08/20/97 at 13:40:07 From: Doctor Rob Subject: Re: Modern Algebra Imaginary numbers are ones whose squares are negative. The simplest one is the square root of -1, which is called "i". Thus i^2 = -1, and i = Sqrt[-1]. Every other imaginary number is a real multiple of i. If j^2 < 0, then j = a*i for some real number a, namely a = Sqrt[-j^2]. Complex numbers are sums of real numbers and imaginary numbers. Often they are written as z = a + b*i, where a and b are real. a is called the real part of z, and b is called the imaginary part of z. (Careful here: b is a real number, but it is still called the imaginary part of z. You might think that b*i, being imaginary, should bear that name, but for historical reasons, that isn't so.) You can think of complex numbers as pairs of real numbers (a,b), and you can plot them in the xy-plane. If you do this with z = 1 - i, you get the point (1,-1) in the fourth quadrant. The imaginary numbers will lie along the y-axis, and the real ones on the x-axis. Now adding complex numbers corresponds to adding the coordinates. Multiplying, (a+b*i)*(c+d*i) = a*c + a*d*i + b*c*i + b*d*i^2 = a*c + a*d*i + b*c*i + b*d*(-1) = (a*c - b*d) + (a*d + b*c)*i. This corresponds to the point (a*c-b*d, a*d+b*c) in the xy-plane. This multiplication may look unfamiliar, but a better idea of it may be gotten by looking at the xy-plane in polar coordinates (r,theta). Recall x = r*cos(theta) and y = r*sin(theta) are the equations for converting from polar coordinates to rectangular ones, and r = Sqrt[x^2+y^2] and theta = arctan(y/x) (with careful attention to the choice of arctangent value) are the equations for converting in the opposite direction. In the case of 1-i, you get r = Sqrt[2] and theta = -Pi/4 + 2*Pi*k, for any integer k. Now take two points (a,b) and (c,d), and write them in polar coordinates: r = Sqrt[a^2+b^2], theta = arctan(b/a), and s = Sqrt[c^2+d^2], phi = arctan(d/c). Then the product of the points is (a*c-b*d, a*d+b*c), where a*c - b*d = r*cos(theta)*s*cos(phi) - r*sin(theta)*s*sin(phi) = r*s*[cos(theta)*cos(phi) - sin(theta)*sin(phi)] = r*s*cos(theta+phi), a*d + b*c = r*cos(theta)*s*sin(phi) + r*sin(theta)*s*cos(phi) = r*s*[cos(theta)*sin(phi) + sin(theta)*cos(phi)] = r*s*sin(theta+phi). Thus multiplying (r,theta) by (s,phi), you get (r*s,theta+phi). That's a bit simpler, and easier to remember. Furthermore, it suggests that theta and phi are actually exponents of something, since multiplying points corresponds to adding these angles. The actual statment of this is called DeMoivre's Law, and it says that x + y*i = r*e^(i*theta) = r*cos(theta) + r*sin(theta)*i. Euler proved that e^(i*Pi) = -1, which agrees with this, so e^(i*2*Pi) = 1. DeMoivre's Law is the key to defining what we might mean by raising a complex number to a complex power. (x + y*i)^(a + b*i) = [r*e^(i*theta)]^(a + b*i) = [e^(ln(r))*e^(i*theta)]^(a + b*i) = e^([ln(r) + i*theta]*[a + b*i]) = e^([a*ln(r) - b*theta] + [b*ln(r) + a*theta]*i) = e^[a*ln(r) - b*theta]*e^([b*ln(r) + a*theta]*i) This gives the polar coordinates of the resulting point. The first coordinate is e^[a*ln(r) - b*theta] and the second is [b*ln(r) + a*theta]. in rectangular coordinates, you get e^[a*ln(r) - b*theta]*cos[b*ln(r) + a*theta] + e^[a*ln(r) - b*theta]*sin[b*ln(r) + a*theta]*i This is rather complicated, but in your case, things are a bit simpler. (1-i)^(2*i) = [Sqrt[2]*e^(-i*Pi/4)]^(2*i) = Sqrt[2]^(2*i)*e^(-2*i^2*Pi/4) = (2^i)*e^(Pi/2) = (2^i)*e^1.570796... but furthermore = e^(Pi/2)*(cos[ln(2)] + sin[ln(2)]*i) = 3.700406 + 3.073709*i, approximately. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/