Friday, 1 August 2014

CHAPTER 1 - COMPLEX NUMBERS - Basic Definitions

It is an amazing fact that by adjoining the imaginary unit i to the real numbers we obtain a complete number field called `` The Complex Numbers." In this amazing number field every algebraic equation in z with complex coefficients
has a solution. To prove this fact we need Liouville's Theorem, but to get started using complex numbers all we need are the following basic rules.

Rules of Complex Arithmetic

  1. displaymath151
  2. Every complex number has the ``Standard Form''displaymath153
    for some real a and b.
  3. For real a and b,displaymath163
  4. displaymath165
  5. displaymath167


Notice that rules 4 and 5 state that we can't get out of the complex numbers by adding (or subtracting) or multiplying two complex numbers together. What about dividing one complex number by another? Is the result another complex number? Let's ask the question in another way. If you are given four real numbers a,b,c and d, can you find two other real numbers x and y so that

OK, so we can divide by c + di if c and d are not both zero. But there is a much easier way to do division.
Notice that

We say that c+di and c-di are complex conjugates. To simplify a complex fraction, multiply the numerator and the denominator by the complex conjugate of the denominator.

Examples of division




Real and Imaginary Parts

If za+bi is a complex number and a and b are real, we say that a is the real part of z and that b is the imaginary part of z and we write

Find tex2html_wrap_inline32 and tex2html_wrap_inline34 .

So that tex2html_wrap_inline36 and tex2html_wrap_inline38

Complex Conjugates

If z=a +bi is a complex number with real part a and imaginary part b, then we denote the complex conjugate of z by tex2html_wrap_inline229 .

Write tex2html_wrap_inline231 in standard form.

so that tex2html_wrap_inline24 .

Prove that tex2html_wrap_inline26 for any integer n.
This follows from the remark made in the solution to the previous problem. Or we can give a simple proof by induction.
We have the result for n=1. Suppose that we know it for integers up to m. We have shown that tex2html_wrap_inline34 for any pair of complex numbers. So we may write


Modulus of a Complex Number

The magnitude or modulus of a complex number z is denoted |z| and defined as

Let zaib where a and b are real so that tex2html_wrap_inline14 . Then

Prove that tex2html_wrap_inline11 .
This is equivalent to tex2html_wrap_inline13 which is

So we just need to show that tex2html_wrap_inline17 . But if za + ib for real numbers a and b then tex2html_wrap_inline25 and tex2html_wrap_inline27 . We could, of course, have verified the identity directly and suggest that the reader do this.

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