# Complex Numbers - A Basic Introduction

Discussion in 'Math' started by Nirvana, Aug 27, 2006.

1. ### Nirvana Thread Starter Well-Known Member

Jan 18, 2005
58
0
In engineering, science and mathematics we are often faced with numbers which can't be evaluated such as
√-64, the answer can't be -8 as -8 x -8 = 64. The answer can't be 8 as 8 x 8 = 64 and so on. So how can we evaluate such a number? Well the answer is to split √-64 into to parts, remember that we can write this number as
√-1 x 64 which is √-1 √64, now we can evaluate the √64 which is 8, but we can't evaluate √-1, so this value of √-1 is given the letter j. (In mathematics √-1 is given the letter i but in electrical engineering we use i to represent current).
Now lets have a look at some powers of j: Well j squared is j x j and we know that as the operations of square and the square root are opposite we simply remove the square operation and we have what is left which in this case is -1. That might sound a bit complicated so I'll use another example which you can work out, take the number √81, now if we square this number we get √81 x √81 which is √81 squared. Now if you work this out you will find that your answer will be what ever number you had in the square root sign which was 81, I hope this is clear.
So back to powers of j then, j x j = √-1 x [FONT=Times New Roman][SIZE=3]√-1 = -1 = j[/SIZE][/FONT]
j x j x j would be j squared x j which is -1 x [FONT=Times New Roman][SIZE=3]√-1 which is [FONT=Times New Roman][SIZE=3]√1. What about j x j x j x j, well this is exactly the same as writing j squared x j squared which is -1 x -1 = 1, so from a completely irrational number of [FONT=Times New Roman][SIZE=3]√-1 we gen get a rational number. So if we were asked to calculate j to the power of nine that is exactly the same as saying j to the four by j to the four by j to the one which is 1 x 1 x [FONT=Times New Roman][SIZE=3]√-1 = [FONT=Times New Roman][SIZE=3]√-1 = j.[/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT]
[FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3]Now lets consider an expression such as 2 + [FONT=Times New Roman][SIZE=3]√-81 , well again in our expression we are faced with this un - solveable [FONT=Times New Roman][SIZE=3]√-81, but we can seperate this out as we did before we can change our expression to; 2 + [FONT=Times New Roman][SIZE=3]√-1 x 81 which is the same as writing; 2 + [FONT=Times New Roman][SIZE=3]√-1 x [FONT=Times New Roman][SIZE=3]√81 , and we can solve [FONT=Times New Roman][SIZE=3]√81 so our expression becomes 2 + ([FONT=Times New Roman][SIZE=3]√-1) x 9 and as we give [FONT=Times New Roman][SIZE=3]√-1 the letter j our expression becomes 2 + j9 also notice that we do have a number in our expression which isn't indeterminate (the number 2) we call these numbers [B]real numbers[/B] and we call the number 9 an [B]imaginary number[/B] as it is associated with the letter j, sometimes called the j operator.[/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT]
[FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3]The next thread will describe complex numbers in greater detail.[/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT]

[FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3][FONT=Times New Roman][SIZE=3]Nirvana.[/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT][/SIZE][/FONT]

2. ### Nirvana Thread Starter Well-Known Member

Jan 18, 2005
58
0
Earlier in the thread titled 'Complex Numbers - An Introduction' the j operator and powers of j were discussed. With this information in hand we will see how we can produce a diagram which represents a complex number. Write the following down on a piece of paper to see where the explanation comes from. Draw a line 3 units long and label the 1 to 3 accordingly, if we multiply the number 3 by -1 we get -3 so you can draw a line in the opposite direction another 3 units long (in the negative x direction). If you can recall from the powers of j the -1 is j squared, therefore instead of saying that we multiplied 3 by -1 we can say that we multiplied it by j squared. Well if we multiplied 3 by j squared j x j , and that rotated our line through 180 degrees, then what angle does our line rotate through if we multiplied it by a single factor of j. So lets write this down, if j x j = 180 the a single j would be 90 as a single j is half that of j x j. So our line that we drew originally which was drawn in the positive x axis direction (if we think of where it would be on a graph), when multiplied by j we rotate through 90 degrees positively (which would be a vertical line), this 90 degrees would be in the position of the y axis on a graph, so we label the positive 90 degrees j and the negative 90 degrees -j this is the axis on which we can now represent our imaginary number and the x axis in the positive and negative directions represent out real number. The diagram looks exactly the same as a normal graph layout which has an x & y axis but the y - axis is replaced by the operator j and -j.

The next thread will deal with writing a complex number in the form r (cosθ + jsinθ).

Nirvana.

3. ### Nirvana Thread Starter Well-Known Member

Jan 18, 2005
58
0
In the following explanation of representing a complex number in the form of r (Cosθ + jSinθ), I would suggest drawing the appropriate diagrams to see where the calculations come from.

Lets start by expressing the complex number 4 + j6, now we are going to draw this on a diagram. Remember that we have an x axis (sometimes called the real axis labeled R) and a j axis, similar to a normal diagram when drawing graphs (called an Argand Diagram).
On the x axis label the units 0 to 4 and on the j (positive) axis label 6 units. Place a point at the appropriate coordinates, that is at point (4,6).
Now from the origin (Point 0 on the diagram) draw a line which extends from the origin (point 0) all the way to your point (4,6). Draw an arrow head on the end at this point to show that is it a vector quantity (has size and direction). From the point (4,6) draw a line (prferably a feint one as it is only for construction purposes, to make it easier to explain what is going on) from the point straight down to the x axis so that we have a vertical line. You should now be able to see that we have a right angled triangle. Now label the line which we drew first from 0 to (4,6) with the letter r, this will represent the magnitude (size) of the line/quantity. Then draw an angle θ from this line ( o to 4,6), to the x axis.
Now the base of our triangle represents our real quantity as the base it the length on the real axis, lets give the base the letter a. Now we will give the vertical line of our triangle (the line which we took from our point (4,6) to the x axis) the letter b. Therefore with reference to the size r and the angle θ, we can describe where our real (letter a ) and our imaginary (letter b) values are. So basically we have a right angled triangle , our hypotenuse is labeled r, our angle θ, the real component a and our imaginary component b. Using Trigonometry we can find our real component a by writing; Cosθ = a/r
so our real component a is found by; a = rCosθ.
(Remember that for now).
Next we can find our imaginary component b by writing; jSinθ = jb/r so our imaginary component is found by;
remember the letter j is there to show that b is an imaginary component, there is a j in front of Sinθ because if the right hand side of our expression is multiplied by j (the jb) the left hand side must also be multiplied by a factor of j (just algebra , one side equals the other). So we have jb = rjSinθ.

Ok remember form the beginning we had a complex number (4 + j6) well this is expressed as (a + jb) and as we worked out from above we can also write this as,
(4 + j6) = ( rCosθ +rjSinθ) take out the common factor of r and we get that (a + jb) = r(Cosθ + jSinθ).
But we havent yet worked out the values of r and θ. Again this is done by Pythagoras and Trigonometry; So r can be found by ; r x r = a x a + b x b (take it to be r squared equals a squared plus b squared), then to find r we take the square root of a squared plus b squared. As a = 4 (the real component) and b = 6 (the imaginary component) then r is equal to the square root of (4 x 4 + 6 x 6) which is the square root of 16 + 36 which is 7.21 . Now what about the angle θ, well we'll work this out from our known (whole numbered) values of a and b. So we get θ = Tan -1 (b/a), θ = Tan -1 (8/4) = 56.31 degrees, (take tan -1 to be the inverse tan function)
Therefore putting it all together we can write our complex number (4 +j6) in the form of (a + jb) which it is and
r(Cosθ + jSinθ), So (a + jb) is (4 + j6) and
r(Cosθ + jSinθ) = 7.21(Cos56.31 + j Sin56.31). There is a shorthanded method of writing this r(Cosθ + jSinθ) just involving r and θ but i'll let you research that one, as I am unable to write it using this writing package.

The next thread will involve the uses of Complex Numbers.

Nirvana.

4. ### gort New Member

Dec 4, 2006
3
0
is the diagram quoted called an argand diagram?, sorry i missed that bit

5. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Yes, the complex space comprised of the real and imaginary planes intersecting orthogonally is known as an Argand Diagram.

If you are interested in learning more about complex numbers and there applications in electronics and circuits, see All About Circuits Volume 2:Chapter 2.

Dave

6. ### alitex Active Member

Mar 5, 2007
122
0
can i share with this?

Complex Numbers
Ali shaban
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Notation

A complex number is written as z = a + jb where a is the real part of z and b is the imaginary part of z. j is the square root of -1 so j2 = - 1. A complex number can also be seen as a vector in a two dimensional space with axes Re z and Im z. In this space the vector will extend from the origin to the point (a, b). Every complex number a + jb has a complex conjugate given by a - jb.
Operations

For 2 complex numbers y = a + jb and z = c + jd the following holds.

y = z only if a = c and b = d. If a complex number is zero, then both its real and imaginary parts are zero.

Addition:- y + z = (a + c) + j(b + d)

Subtraction:- y - z = (a - c) + j(b - d)

Multiplication:- yz = (a + jb) (c + jd) = ac + jad + jbc + j2bd = (ac - bd) + j(ad + bc)

Reciprocal:- 1/(a + jb) = (a - jb)/ [(a + jb)(a - jb)] = (a - jb)/(a2 + b2)

Division:- y/z = (a + jb)/(c + jd) = [(a + jb)(c - jd)]/[(c + jd)(c - jd)] = [(ac + bd) + j(bc - ad)]/(c2 + d2)
This can also be written :- y/z = (ac + bd)/(c2 + d2) + j(bc - ad)/(c2 + d2)

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Polar Form of Complex Numbers

A complex number z = a + jb can be written is polar form as z = r e jq where r2 = a2 + b2. r is called the magnitude or modulus or absolute value of z. and q is the phase or argument of z.

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De Moivre's equation
( a + jb )n = (r ejq )n = rn (ejq )n = rn (ejnq ) = rn ( cos nq + j sin nq)
where r2 = a2 + b2
and q = tan-1(b/a)

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Operations on complex numbers in polar form
Operations like multiplication, division, powers and roots are very easy if complex numbers are in polar form. Consider 3 complex numbers:-
z = r e jq
z1 = r1 e jq1
z2 = r2 e jq2

Multiplication:- z1z2 = r1r2 e j(q1 + q2)

Division:- z1/z2 = r1/r2 e j(q1 - q2)

Powers:- zn = rn e jnq

Roots:- z1/n = r1/n e j(q + 2mpi)/n

Logs:- ln z = ln r + j(q + 2mpi)

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nth roots of unity
11/n = [e j 2pm]1/n = cos(2pm/n) + j sin(2pm/n)
The nth roots of unity are obtained for values of m from 0 to n - 1.

(-1)1/n = [e j (2m + 1)p]1/n = cos(p(2m + 1)/n) + j sin(p(2m + 1)/n)
The nth roots of -1 are obtained for values of m from 0 to n - 1.

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Complex Numbers in Oscillations and Waves
For oscillations, the following relation is useful. e jw t = cos w t + j sin w t
cos w t = Re [e jw t]
sin w t = Im [e jw t]
For waves, the following is useful. e j(kx - w t) = cos(kx - w t) + j sin(kx - w t)

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Functions of a Complex Variable
A function f(s) = 1/(s + a), where s is a complex number, is a function of a complex variable s or a complex function. Of particular interest in complex functions is the location of the poles(points at which the function goes to infinity) of the function. In f(s) there is only one pole at s = - a.

7. ### thenzman New Member

May 17, 2007
5
0
What is the use of De Moivre's equation? I know how to do it and I did it for the test but I don't know why am doing it. What are we trying to find when using De Moivre's equation?

THANKS,

Feb 24, 2006
10,338
1,850
9. ### thenzman New Member

May 17, 2007
5
0
THANKS papabravo the wikipedia link was helpful in a way. I have some idea of what it is used for. Is there any practical example you could give?

10. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
In practical examples, it's mainly used as an identity that can make some calculations easier to manipulate. Given that versions of the identity have exp(i*x) you can quite feasibly see it being used in Fourier analysis, which has applications in many fields including Digital Signal Processing, Communication Engineering and Image Processing.

Dave

11. ### gbm46 Active Member

May 6, 2007
46
0
This is a paradox I found interesting...

i = i
sqrt( -1 ) = sqrt( -1 )
sqrt( 1 / -1) = sqrt( - 1 / 1 )
sqrt( 1 ) / sqrt( -1 ) = sqrt (-1 ) / sqrt ( 1 )
sqrt ( 1 ) * sqrt( 1 ) = sqrt ( -1 ) * sqrt( -1 ) (cross-multiplying)
1 = -1

This is not a real paradox because i cannot be treated like a 'real' square root. Lesson: you cant do this with i (see this has a point). Similarly 1 = 2 easily enough if we are allowed to divide by zero.

12. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
If we are allowed to divide by zero and to cross multiply with i, would -1 = 2?

13. ### recca02 Senior Member

Apr 2, 2007
1,211
0
similar operation that wud violate the rule of maths is
sqrt(-3)*sqrt(-3) = sqrt(3)*i*sqrt(3)*i
sqrt(-3*-3) = i^2*sqrt(3)^2
sqrt(9) = -1*3
3=-3
once 3 = -3 is proven by simple multiplication and division anything like 4563=786 can easily be proven
the mistake above is obvious normal radical operations like those can not be done on a complex number they mus always be first converted into X*i and solved.

14. ### m4yh3m Senior Member

Apr 28, 2004
186
42
Who knew complex numbers were so... complex? They should be called "OMGWTFBBQ" numbers.

15. ### mt340179 New Member

Nov 14, 2007
1
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Curiosity: Why do we have to allow the real and imaginary plane to intersect orthogonally?? Are there not some properties that would benefit in letting the planes intersect in some radian proportion?

16. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Its because the Argand Diagram adopts the Cartesian coordinate system to describe what is essentially a two-coordinate representation; i.e. real and imaginary components.

Can you give us an example of where you think there would be a benefit in letting the planes intersect in some radian proportion?

Dave

17. ### chesart1 Senior Member

Jan 23, 2006
269
1
If you check your text books, you will find that i^2 = -1. The book Introductory Circuit Analysis written by Robert Boylestad defines j^2 (where j is the square root of -1) in chapter 14 titled Phasors.
Hence, the term i (in your manipulation) can be treated as a real term when it is being squared.

1/j = 1(1/j) = (j/j)(1/j) = j/(j^2) = j/(-1) = -j

The square root of 1 is 1. Therfore ...

i = i
sqrt( -1 ) = sqrt( -1 )
sqrt( 1 / -1) = sqrt( - 1 / 1 ) error: 1/j = -j/1
sqrt( 1 ) / sqrt( -1 ) = sqrt (-1 ) / sqrt ( 1 )
sqrt ( 1 ) * sqrt( 1 ) = sqrt ( -1 ) * sqrt( -1 ) (cross-multiplying)
1 = -1