Download Complex numbers 1

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Complex numbers 1.
You have previously been taught that the solution to quadratic equations
like ax2 + bx + c = 0 is:
x = (-b   b2 – 4 a c )/2a
Example 1
Now, solve the following quadratic equation:
2x2 + 9x +7 = 0
The solution is:
x = (-9   92 – 4 * 2 *7 )/2 * 2 = (-9   25)/4
The results are that x is either –1 or –3.5
Now, try this one:
Example 2
5x2 – 6x + 5 = 0
If you try to solve this in the same way as you did the first one you get:
x = (6   62 – 4 *5 *5 )/2 * 5 = (6   -64)/10
Now you have to find what the square root of (-64 ) is. It can not be 8
and it can not be –8 either as both these give the value +64 when they are
squared. So what now? The truth is that  -64 simply can not be
represented by an ordinary number. The reason for this is that there is
no real number that becomes a negative number when it is squared. So
where does this lead us? Well, -64 could be considered to be (64) * (-1)
and that gives us  (-1) (64) which is 8 *  (-1). Of course we are faced
with the same problem when it comes to evaluating  (-1) for the same
reason as given above. So now we define a new number. This number is
called j and is defined as j2 = -1 so that j itself =  (-1). Now the equation
we started above can be completed. The result is:
x = (6  j8)/10 this give x1 = 0.6 + j0.8 and x2 = 0.6 – j0.8
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Numbers of this kind are called complex numbers. They have one real and
one imaginary part. In example 2 the real part of the complex number is
0.6 and the imaginary part is j 8 * (0.8). We will look a lot more about this
later. Now we need to concentrate on the number j
Since the definition of j is  (-1), we need to consider some powers of j
j =  (-1)
j2 =  (-1) *  (-1) = -1
j3 =  (-1) *  (-1) *  (-1) = -j
j4 =  (-1) *  (-1) *  (-1) *  (-1) = (-1) * (-1) = 1
The last result, j4 is particularly important as it shows that each time a
factor, j4 occurs, it can be replaced by 1 so that the power of j is reduced
to one of the above results. For example:
j9 = (j4)2 *j1 = (1)2 * j = 1* j = j
Another example of this is:
Example 3
J20 = (j4)
5
= (1)5 = 1
Now when you have become familiar with the number j and its behaviour,
we can return to actual complex numbers. You saw from the quadratic
equation in Example 2 that the solution was :
x1 = 0.6 + j0.8 and x2 = 0.6 – j0.8
Here 0.6 is the real part and 0.8 is the imaginary part. {Note that it is
0.8, not j0.8 that is the imaginary part of the complex numbers!!!}
Addition and subtraction of complex numbers
Complex numbers have several applications in engineering. Therefore we
need to be able to carry out the usual arithmetic operations. The first we
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shall look at is addition. The real and imaginary parts can not be added
together. So what we do is to add the real parts and the imaginary parts
separately. In general:
(a + jb) + (c + jd) = (a + c) + j(b + d)
Example 4
(4 + j5) + (3 + j2) = (4 + 3) + j(5 + 2) = 7 + j7
Subtraction is done in the same way. In general we have:
(a + jb) - (c + jd) = (a + c) + j(b - d)
Example 5
(4 + j5) - (3 + j2) = (4 - 3) - j(5 - 2) = 1 – j3
This can be combined:
Example 6
(6 + j5) – (4 – j3) + (2 – j7) = 6 + j5 – 4 + j3 + 2 – j7 =
(6 – 4 + 2) + j(5 + 3 –7) = 4 + j