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.
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.
Find
and 
.
So that
and 
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
.
Exercise:
Write
in standard form.
so that
.
Prove that
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
for any pair of complex numbers. So we may write
Let z= a+ ib where a and b are real so that
. Then
Prove that
.
This is equivalent to
which is
So we just need to show that
. But if z= a + ib for real numbers a and b then 
and 
. We could, of course, have verified the identity directly and suggest that the reader do this.
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
- Every complex number has the ``Standard Form''
for some real a and b. - For real a and b,
Division
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 thatOK, 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 z= a+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 writeFind
and .So that
and 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
.Exercise:
Write
in standard form.so that
.Prove that
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
for any pair of complex numbers. So we may writeModulus of a Complex Number
The magnitude or modulus of a complex number z is denoted |z| and defined asLet z= a+ ib where a and b are real so that
. ThenProve that
.This is equivalent to
which isSo we just need to show that
. But if z= a + ib for real numbers a and b then and . We could, of course, have verified the identity directly and suggest that the reader do this.
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