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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

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.

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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