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Plotting Complex Numbers


Date: 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/   
    
Associated Topics:
College Imaginary/Complex Numbers
College Modern Algebra

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