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| An introduction to complex numbers |
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If we desire that every integer has an inverse element, we have to invent rational numbers and many things become much simpler. If we desire every polynomial equation to have a root, we have to extend the real number field R to a larger field C of 'complex numbers', and many statements become more homogeneous. To construct a complex number, we associate with each real number a second real number. A complex number is then an ordered pair of real numbers (a,b). We write that new number as a + bi The '+' and the i are just symbols for now. We call 'a' the real part and 'bi' the imaginary part of the complex number. Ex : (2 , 4.6) or 2 + 4.6i ;
(0 , 5) or 0 + 5i ;
(-5 , 36/7) or -5 + (36/7)i ;
Instead of 0 + bi, we write 5i.
Instead of a + 0i, we write a.
Instead of 0 + 1i, we write i.
The set of all complex numbers is C. A complex number has a representation in a plane. Simply take an x-axis and an y-axis (orthonormal) and give the complex number a + bi the representation-point P with coordinates (a,b). The point P is the image-point of the complex number (a,b).  The plane with all the representations of the complex numbers is called the Gauss-plane. With the complex number a + bi corresponds just one vector OP or P. The image points of the real numbers 'a' are on the x-axis. Therefore we say that the x-axis is the real axis. The image points of the 'pure imaginary numbers' 'bi' are on the y-axis. Therefore we say that the y-axis is the imaginary axis. Two complex numbers (a,b) and (c,d) are equal if and only if (a = c and b = d).
So a + bi = c + di
or a = c and b = d
We define the sum of complex numbers in a trivial way.
(a,b) + (a',b') = (a + a',b + b')
or (a + bi) + (a'+ b'i) = (a + a') + (b + b')i Ex. (2 + 3i) + (4 + 5i) = 6 + 8i If (a + bi) corresponds with vector P in the Gauss-plane and (a' + b'i) corresponds with vector P', then we have : co( P)=(a,b) and co( P')=(a',b') => co( P + P')=(a,b) + (a',b') => co( P + P')=(a + a',b + b') So P + P' is the vector corresponding with the sum of the two complex numbers. The addition of complex numbers correspond with the addition of the corresponding vectors in the Gauss-plane.  We define the product of complex numbers in a strange way. (a,b).(c,d)=(ac - bd,ad + bc) Ex. : (2 + 3i).(1 + 2i)=(-4 + 7i) Later on we shall give a geometric interpretation of the multiplication of complex numbers. The importance of that strange product is connected with (0,1).(0,1)=(-1,0) or the equivalent
Here we see the importance of that strange definition of the product of complex numbers. The real negative number -1 has i as square root! We write a + 0i as a. We write 0 + 1i as i. a . i = (a + 0i)(0 + 1i) = (0 + ai) = ai Therefore, the product a . i is the same as the notation ai. We write a + 0i as a. We write 0 + bi as bi. So (a) + (bi) = (a + 0i) + (0 + bi) = a + bi Therefore, the sum of a and bi is the same as the notation a + bi Because (a + bi)+((-a) + (-b)i)=0 + 0i , we call (-a) + (-b)i the opposite of a + bi. We write this opposite of (a + bi) as -(a + bi). So, the opposite of bi is (-b)i = -bi We define (a + bi) - (c + di) as (a + bi) + (-c + (-d)i). So, (a + bi) - (c + di)=((a - c) + (b - d)i and a + (-b)i=a - bi We define the conjugate of a + bi as a + (-b)i = a - bi
Notation: a + bi = conj(a + bi) = a - bi Ex : 2 + 3i = 2 - 3i We define modulus or absolute value of a + bi as sqrt(a 2 + b 2) . We write this modulus of a + bi as |a + bi|. If p is the representation of a + bi in the Gauss-plane, the distance from O to P is the modulus of a + bi. Ex: |3 + 4i| = 5 It can be proved that C, + , . is a field. Therefore we can apply all the properties of a field to the calculation with complex numbers. We apply the law of distributivity to (a + bi).(c + di)and note that i.i=-1 (a + bi).(c + di) = ac + adi + bci - bd = (ac - bd) + (ad + bc)i To divide (a + bi) by (c + di), we multiply the numerator and the denominator with the complex conjugate of the denominator. a + bi) (a + bi)(c - di)
-------- = -------------------
(c + di) ((c + di)(c - di))
(ac + bd) + i(bc - ad)
= -----------------------
(c2 + d2)
(ac + bd) (bc - ad)
= --------- + i-----------
(c2 + d2) (c2 + d2)
The only square root of 0 is 0.
If a is a strict positive real number, we know that a has two real square roots. It can be proved that there are no other square roots of a in C. As b is a negative real number, -b is strict positive and has two square roots c and -c. So -b = c2 = (-c)2 and b = (ic)2 = (-ic)2 i. sqrt(-b) and -i. sqrt(-b) are the square roots of the negative real number b. Ex : 3i and -3i are the square roots of -9. It can be proved that there are no other square roots of b in C. We are looking for all real numbers x and y so that (x + iy)(x + iy) = a + ib (1) <=> x2 - y2 + 2xyi = a + bi (2) <=> x2 - y2 = a and 2xy = b (3) Because b is not 0, y is not 0 and so <=> x2 - y2 = a and x = b/(2y) b2 corresponding x-value. Let t = y2 in the first equation of (4) then 4t2 + 4at - b2 = 0 (5) Let r = modulus of a + bi The discriminant = 16(a 2+ b2) = 16r2 We note the roots as t1 and t2. <=> t1 = (- a + r)/2 and t2 = (- a - r)/2 (6) Since y is real and r > a, t 1 > 0 and gives us values of y. Since the product of the roots of (5) is (-b 2/4) < 0 , t 2 is strictly negative. So we find two values of y. We note these values y 1 and y 2. y1 = sqrt((r - a)/2) and y2 = -sqrt((r - a)/2) (7) The corresponding x values are x1 = b/(2.y1) and x2 = b/(2.y2) (8) Note that the two solutions are opposite complex numbers. So any (not real) complex number has two opposite complex roots. They can be calculated with the formulas (7) an (8). The two square roots of a+bi are (x +yi) and -(x +yi) with
y = sqrt((r - a)/2) and x = b/(2.y) Ex1. We calculate the square roots of 3 + 4i. |3 + 4i| = 5 ; y = sqrt((5 -3)/2) = 1 and x = 4/2 = 2 The square roots of 3 + 4i are 2 + i and -2 - i Ex2. We calculate the square roots of 6 + 8i |6 + 8i| = 10 ; y = sqrt((10 - 6)/2) = sqrt(2) and x = 8/(2 sqrt(2)) = 2 sqrt(2) The square roots of 6 + 8i are 2 sqrt(2) + sqrt(2)i) and -(2 sqrt(2) + sqrt(2)i) We already know that r = sqrt(a 2 + b 2) is the modulus of a + bi and that the point p(a,b) in the Gauss-plane is a representation of a + bi. The intersection point s of [op and the goniometric circle is s( cos(t) , sin(t) ). That number t, a number of radians, is called an argument of a + bi. We say an argument because, if t is an argument so t + 2.k.pi is an argument too. Here and in all such expressions k is an integer value. We just saw that s( cos(t) , sin(t) )
and we have the vector-equation os = r. op
Therefore p( rcos(t) , rsin(t) ) but also p(a,b).
It follows that a = rcos(t) ; b = rsin(t).
So we have a + ib = rcos(t) + i rsin(t) or a + ib = r (cos(t) + i sin(t))
r(cos(t) + i sin(t)) is called the polar representation of a+bi. If z and z' are complex numbers, they have the same representation in the Gauss-plane. So they have the same modulus and the arguments difference is 2.k.pi We have : r(cos(t) + i sin(t)) = r'(cos(t') + i sin(t')) <=> r = r' and t = t' + 2.k.pi We have : r(cos(t) + i sin(t)).r'(cos(t') + i sin(t')) = rr'(cos(t).cos(t') - sin(t)sin(t') + i cos(t)sin(t') +i sin(t)cos(t')) = rr'(cos(t+t') + i sin(t + t')) Rule: To multiply two complex numbers, we multiply the moduli and add the arguments. This rule can be extended to the multiplication of n complex numbers. r(cos(t) + i sin(t)). r'(cos(t') + i sin(t')) = r r'(cos(t+t') + i sin(t + t')) With this rule we have a geometric interpretation of the multiplication of complex numbers. The image-points of conjugate complex numbers lie symmetric with regard to the x-axis. So conjugate complex numbers have the same modulus and opposite arguments. The conjugate of r(cos(t) + i sin(t)) is r(cos(-t) + i sin(-t)) We have : r(cos(t) - i sin(t))/r'(cos(t') + i sin(t'))= r(cos(t) + i sin(t)) / r'(cos(t') + i sin(t'))R = r(cos(t) + i sin(t)) (cos(-t') + i sin(-t')) / r' = r .(cos(t - t') + i sin(t - t')/r' Rule: To invert a complex number, we invert the modulus and we take the opposite of the argument. 1 / r(cos(t) + i sin(t)) = (1/r)(cos(-t) + i sin(-t))
Using the multiplication rule we have:
( r (cos(t) + i sin(t)) )n = rn .(cos(nt) + i sin(nt))
Take the previous formula with r = 1. (cos(t) + i sin(t))n = cos(nt) + i sin(nt) First take the inverse and then the positive power. (r(cos(t) + i sin(t)))n = (1/r)n (cos(-t) + i sin(-t))n = (1/r)n (cos(-nt) + i sin(-nt)) Take a complex number c = r(cos(t) + i sin(t)). We are looking for all the complex numbers c' = r'(cos(t') + i sin(t')) so that (c')n = c <=> (r')n (cos(nt') + i sin(nt')) = r(cos(t) + i sin(t)) <=> (r')n = r and nt' = t + 2k.pi <=> r'= positive nth-root-of r and t' = (t/n) + k.(2pi/n) If r and t are known values, it is easy to calculate r' and different values of t'. Plotting these results in the Gauss-plane, we see that there are just n different roots. The image-points of these numbers are the angular points of a regular polygon. A complex number c = r(cos(t) + i sin(t)) has exactly n n-th roots. r'(cos(t') + i sin(t')) is a complex n-th root of c if and only if r'= positive nth-root-of r and t' = (t/n) + k.(2pi/n)
with k = 0, ... ,n-1
Ex.
We calculate the 6-th roots of (-32 + 32.sqrt(3).i)
The modulus is r = 64. The argument is (2.pi/3).
r' = 2 and t' = pi/9 + k.(2pi/6)
The roots are 2(cos(pi/9 + k.(2pi/6)) + i sin(pi/9 + k.(2pi/6))) for (k = 0,1,..,5)
In the Gauss-plane these are the angular points of a regular hexagon.
It can be shown that many formulas and properties for polynomials with real coefficients also hold for polynomials with complex coefficients. Examples: - The formulas to solve a quadratic equation.
- The formulas for sum and product of the roots of a quadratic equation.
- The methods for dividing polynomials
- Horner's method
- Methods to factor a polynomial
- Properties about divisibility
- ...
Each polynomial equation with complex coefficients and with a degree n > 0, has at least one root in C. Each polynomial equation with complex coefficients and with a degree n > 0, has exactly n roots in C.
These roots are not necessarily different.
Proof:
We give the proof for n=3, but the method is general.
Let P(x)=0 the equation.
With d'Alembert we say that P(x)=0 has at least one root b in C.
Hence P(x)=0 <=> (x-b)Q(x)=0 with Q(x) of degree 2.
With d'Alembert we say that Q(x)=0 has at least one root c in C.
Hence P(x)=0 <=> (x-b)(x-c)Q'(x)=0 with Q'(x) of degree 1.
With d'Alembert we say that Q'(x)=0 has at least one root d in C.
Hence P(x)=0 <=> (x-b)(x-c)(x-d)Q"(x)=0 with Q"(x) of degree 0. Q"(x)=a .
Hence P(x)=0 <=> a(x-b)(x-c)(x-d)=0 .
From this, it follows that P(x)=0 has exactly 3 roots.
If c is a root of a polynomial equation with real coefficients, then conj(c) is a root too.
We give the proof for n=3, but the method is general.
P(x) = a x3 + b x2 + d x + e Since c is a root of P(x) = 0 , we have a c3 + b c2 + d c + e = 0 => conj(a c3 + b c2 + d c + e)= 0 => a conj(c)3 + b conj(c)2 + d conj(c)+ e = 0 => conj(c) is a root of P(x) = 0. Now, we know that if c is a root of a polynomial equation with real coefficients, then conj(c) is a root too.
The roots that are not real, can be gathered in pairs, c and conj(c).
Then,the polynomial P(x) is divisible by (x-c)(x-conj(c)) and this is a real quadratic factor.
So each polynomial with real coefficients can be factored into real linear factors and real quadratic factors.
Say P(x) = a x5 + bx4 + cx3 + dx2 + ex + f We factor this polynomial: P(x) = a(x-g)(x-h)(x-i)(x-j)(x-k) Then P(x) = a(x5 - (g+h+i+j+k)x4 + ...+(-1)5 ghijk) Hence , -a(g+h+i+j+k) = b and a((-1)5 ghijk) = f The sum of the roots is -b/a This formula holds for every polynomial equation ! The product of the roots is (-1)5 f/a For a polynomial equation of degree n, we have The product of the roots is (-1)n l/a . (l is the last coefficient)
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(posted on 6 Sep 2007 22:21:54 IST)
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| my i revised my complexes reading this bravo man |
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(posted on 7 Sep 2007 09:18:24 IST)
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THANKXXX SIR, IT HELPED ME A LOT!!!!!!!!
HATS OFF TO UR. SINCERE EFFORTS............. |
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