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.
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.
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a + bi = c + di |
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(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
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i.i = -1 or i2 = -1. |
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
(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
------
a + bi = conj(a + bi) = a - bi
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______
Ex : 2 + 3i = 2 - 3i
We define
modulus or absolute value of a + bi as sqrt(a2 + b2) .
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
We apply the law of distributivity to (a + bi).(c + di)and note that i.i=-1
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(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)
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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
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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 b
<=> ---- - y2 = a and x = ---- (4)
4y2 2y
The first equation of (4) gives us y and the second gives the 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, t1 > 0 and gives us values of y.
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).
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The two square roots of a+bi are (x +yi) and -(x +yi) with |
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 - iEx2. 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(a2 + b2) 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.
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a + ib = r (cos(t) + i sin(t)) |
r(cos(t) + i sin(t)) is called the polar representation of a+bi.
r(cos(t) + i sin(t)) = r'(cos(t') + i sin(t'))
<=> r = r' and t = t' + 2.k.pi
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.
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r(cos(t) + i sin(t)). r'(cos(t') + i sin(t')) = r r'(cos(t+t') + i sin(t + t')) |
| The conjugate of r(cos(t) + i sin(t)) is r(cos(-t) + i sin(-t)) |
1 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(cos(t) - i sin(t)) 1
= ---------------------- = -.(cos(-t) + i sin(-t))
r2 r
Rule: To invert a complex number, we invert the modulus and we take the opposite of the argument.
1 1
-------------------- = -.(cos(-t) + i sin(-t))
r(cos(t) + i sin(t)) r
|
r(cos(t) + i sin(t)) 1
----------------------- = r(cos(t) + i sin(t)).----------------------
r'(cos(t') + i sin(t')) r'(cos(t') + i sin(t'))
1
= r(cos(t) + i sin(t)) . ---(cos(-t') + i sin(-t'))
r'
r
= - .(cos(t - t') + i sin(t - t')
r'
Rule: To divide two complex numbers, we divide the moduli and subtract the arguments.
With this rule we have a geometric interpretation of the division of complex numbers.
r(cos(t) + i sin(t)) r ----------------------- = - .(cos(t - t') + i sin(t - t') r'(cos(t') + i sin(t')) r' |
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( r (cos(t) + i sin(t)) )n = rn .(cos(nt) + i sin(nt)) |
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(cos(t) + i sin(t))n = cos(nt) + i sin(nt) |
(r(cos(t) + i sin(t)))n = (1/r)n (cos(-t) + i sin(-t))n = (1/r)n (cos(-nt) + i sin(-nt))
Say c and c' are two complex numbers.
We write conj(c) for the conjugate of c.
We write mod(c) for modulus of c.
With previous formulas it is easy to prove that
conj(c.c') = conj(c).conj(c') (extendable for n factors) conj(c/c') = conj(c)/conj(c') conj(c + c') = conj(c) + conj(c') mod(c.c') = mod(c).mod(c') (extendable for n factors) mod(c/c') = mod(c)/mod(c') |
(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.
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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) |
| Each polynomial equation with complex coefficients and with a degree n > 0, has exactly n roots in C. |
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. |
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.
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)
You can find solved problems about complex numbers
here