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Complex Numbers
Most of the familiar mathematical operations that we perform on numbers are closed and
complete: add together any two numbers and you get another number; multiply together any two
numbers and you get another number, and so on. You can never get a result that isn’t a number,
and (ignoring the case of division by zero) all numbers can be inputs to these operations. It
makes arithmetic seem nice and solid and useful.
But square-root is different. Negative numbers don’t have square roots, or at least if they do,
their square roots are not really numbers.
That can cause trouble. For example, there might be a formula for the required amount of some
substance looking something like this:
Required amount (kg)
= 3(k-1)2+22
Where k is chosen to satisfy k2-2k+5 = 0
You know how to solve that sort of problem: first find out what k must be, using the quadratic
equation, then use k to find out the required amount.
Ax2 + Bx + C = 0  x = -B  sqrt(B2 – 4AC)
2A
so we have A=1, B=-2, C=5, and k = ( 2  sqrt((-2)2 – 415) ) / (21)
= ( 2  sqrt(4 - 20 ) / 2
= ( 2  sqrt(-16) / 2
oh dear, the value of k has the square root of a negative number in it, so it doesn’t exist, and we
can’t find out what the required amount is.
That seems like the end of it, the problem can not be solved. Except for the fact that there is a
perfectly valid real numeric solution, it is just that we can’t find it.
This sort of thing does come up surprisingly frequently in mathematics and engineering: you
need the square root of a negative number on the way to finding a real solution to a real problem.
The way to do it is simply to pretend that negative numbers do have square roots.
Of course, we can’t just pick on a normal number and pretend that it is the square root of minus
sixteen. That would stop all of mathematics from working. No existing number can possibly be
the square root of minus one, so instead we use something that isn’t a number. Something that
has no existing mathematical meaning that could be ruined.
We define a new thing, not a number, called , by stating 2 = -1
Do not be concerned that no such number exists, it is not supposed to be a real number.
With that definition,  is a perfectly good value for -1, and so of course is -.
And if -1 has a square root, then so does every other negative number. the square root of minus
four must be 2; the square root of minus forty-nine must be 7; the square root of minus two
must be 1.41421356, and so on.
These new number-like things that are multiples of , and stand in for the square roots of
negative numbers, are called Imaginary Numbers, to distinguish them from the Real Numbers
that we are familiar with. They really are quite easy to deal with.
But, after adding these new kinds of numbers, we haven’t got closure any more. Some operations
are OK, multiplication for instance:
if you multiply together two real numbers, you get a real number as always: 23=6;
if you multiply two imaginary numbers, you get a real number: 23 = 62 = 6-1 = -6;
if you multiply an imaginary number by a real number, you get an imaginary number: 23=6.
Addition nearly works:
if you add together two real numbers, you get a real number as always: 2+3=6;
if you add two imaginary numbers, you get an imaginary number: 2+3 = 6;
BUT if you add an imaginary number to a real number, what do you get?
2+3 can be neither real nor imaginary, it must be a combination of the two.
So now we have to invent a third kind of number. A Complex Number is defined to be a Real
number and an Imaginary number added together. For example, 2+3.
Surprisingly, this new invention simplifies the whole situation. We now only ever need to deal
with one kind of number, because both real and imaginary numbers are completely covered by
complex numbers. The real number 7 can be written as the complex number 7+0; the imaginary
number 3 can be written as the complex number 0+3.
The whole system of complex numbers is closed and complete.
Add together two complex numbers, and the result is a complex number;
Subtract one complex number from another, and the result is a complex number;
Multiply together two complex numbers, and the result is a complex number;
Divide one complex number by another (except 0+0), and the result is a complex number;
Even complex numbers have complex numbers as their square roots.
Let’s go back to the original problem, and solve it:
Required amount (kg)
= 3(k-1)2+22
Where k is chosen to satisfy k2-2k+5 = 0
we had A=1, B=-2, C=5, and k = ( 2  sqrt((-2)2 – 415) ) / (21)
= ( 2  sqrt(4 - 20 ) / 2
= ( 2  sqrt(-16) / 2
now continuing...
= ( 2  4 ) / 2
k = 1+2 or 1-2
so we take one of our solutions, and put it into the formula for “required amount”:
Required amount (kg)
= 3(k-1)2+22
=
=
=
=
=
3(1+2-1)2+22
3(2)2+22
3-4+22
-12+22
10
Even though the computation required the use of non-existent complex numbers, we still got a
perfectly reasonable real number as the result. Problem Solved.
Working out how to do arithmetic.
We’ll work on generic complex numbers (A+B) and (C+D), to work out general rules, instead
of trying to solve specific examples.
Addition is easy: (A+B) + (C+D) is simply (A+C) + (B+D), just add the real parts and the
complex parts separately.
Example: (2+4) + (5-2) = 7 + 2
Subtraction is exactly the same: (A+B) - (C+D) is simply (A-C) + (B-D), just subtract the
real parts and the complex parts separately.
Example: (2+4) - (5-2) = -3 + 6
For multiplication, we need to expand the product out, as when doing algebra:
(A+B)  (C+D) = A(C+D) + B(C+D)
= AC + AD + BC + BD
= AC + AD + BC + BD
= AC + (AD + BC) + BD-1
= (AC - BD) + (AD + BC)
Example: (2+4)  (5-2) = (25 - 4-2) + (2-2 + 45)
= (10 - -8) + (-4 + 20)
= 18 + 16
Division takes a little trick. Anything divided by itself is 1, so we can multiply the whole formula
by (C-D)  (C-D) without having any effect:
(A+B)  (C+D) = ((A+B)  (C+D))  ((C-D)  (C-D))
= ((A+B)  (C-D))  ((C+D)  (C-D))
= (A(C-D) + B(C-D))  (C(C-D) + D(C-D))
= (AC-AD + BC-BD)  (CC-CD) + DC-DD)
= (AC-AD + BC-BD2)  (C2-CD + CD-D22)
= (AC-AD + BC +BD)  (C2-CD + CD +D2)
= ((AC+BD) +(BC - AD) )  (C2 +D2)
= ((AC+BD)(C2+D2)) + ((BC-AD)(C2+D2))
Example: (2+4)  (5-2)
first calculate (C2+D2) = (52+(-2)2) = 29
(AC+BD) = (25+4(-2)) = 2
(BC-AD) = (45+2(-2)) = 16
so (2+4)  (5-2) = (229) + (1629)
= 0.069 + 0.552
One final useful operation is the Magnitude of a complex number: how “big” it is, or in other
words, how far away from zero is it. Calculated from Pythagoras’ theorem:
Magnitude( A+B ) = sqrt(A2 + B2)
Example: magnitude(5-2) = sqrt(52+(-2)2) = sqrt(29) = 5.385