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 |
|
(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
|
______
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
| (a + bi).(c + di) = ac + adi + bci - bd = (ac - bd) + (ad + bc)i |
(2+3i). (5 - 2 i) = 16 + 11i
i.(3-i).i = -3 + 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)
|
16+11i ------ = 2 + 3i 5 - 2i 15-7i ----- = -7 - 15 i i
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.
i and -i are the square roots of -1.
a.i and -ai are the square roots of -a2.
a.b.i and -a.b.i are the square roots of - a2b2.
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 There is a second method to calculate the two complex roots. A little further on this page, this second method is shown. |
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.
We write this modulus as |a+bi|.
We also know that the point p(a,b) in the Gauss-plane is a representation of a + bi.
The intersection point s of [op and the trigonometric 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.
Examples:
i = 1(cos(pi/2) + i sin(pi/2))
1+i = sqrt(2).( cos(pi/4) + i sin(pi/4) )
3+4i = 5 ( cos(0.927295218002) +i sin(0.927295218002) )
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.
Exercise:
Draw the complex numbers c1=1+i and c2=2-i in the Gauss-plane.
Then draw the product c3=3+i.
Measure the modulus of each of those numbers and see that |c1|.|c2| = |c3|.
Measure the argument of each of those numbers and see that
argument of c1 + argument of c2 = argument of c3.
Example:
5 ( cos(1.2) + i sin(1.2) ) * 4 ( cos(-0.2) + i sin(-0.2) ) = 20 (cos(1)+ i sin(1))
|
r(cos(t) + i sin(t)). r'(cos(t') + i sin(t')) = r r'(cos(t+t') + i sin(t + t')) |
Draw a complex number c (vector) in the Gauss-plane. Multiply c by i. Determine that the result arises by rotating c, with center O, over a right angle.
Draw two complex numbers with modulus 1 and construct the product.
| The conjugate of r(cos(t) + i sin(t)) is r(cos(-t) + i sin(-t)) |
Draw two conjugate complex numbers. Construct the product. Note that the product is a real number.
Prove that the product of two complex numbers als always a real number.
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
|
Draw a complex number and its inverse in the Gauss-plane.
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' |
Example:
5 ( cos(1.2) + i sin(1.2) ) ---------------------------- = 1.25 ( cos(1.4) + i sin(1.4) ) 4 ( cos(-0.2) + i sin(-0.2))
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( r (cos(t) + i sin(t)) )n = rn .(cos(nt) + i sin(nt)) |
Example: ( 2( cos1.2 + i sin1.2 ) )5 = 32( cos 6 + i sin 6 )
Find (1+i)5 ; (transform first 1+i to the trigonometric form)
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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))Example : (cos(0.1) + i sin(0.1))-10 = (cos(-1) + i sin(-1)) = (cos(1) - i sin(1))
Say c and c' are two complex numbers.
We write conj(c) for the conjugate 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') |c.c'| = |c|.|c'| (extendable for n factors) |c/c'| = |c|/|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 + 2 k pi/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. Obviously, the two square roots of a complex number can be calculated with this method. |
Example 1
We calculate the square roots of (-32 + 32.sqrt(3).i)
The modulus is r = 64. The argument is (2.pi/3).
The square roots are
8(cos(pi/3) + i sin(pi/3)) and -8(cos(pi/3) + i sin(pi/3))
Example 2
We calculate the 6-th roots of (-32 + 32.sqrt(3).i)
The modulus is r = 64. The argument is (2.pi/3).
The roots are 2( cos(pi/9 + 2 k pi/6) + i sin(pi/9 + 2 k pi/6) ) with k = 0,1,..,5
In the Gauss-plane these roots are the vertices of a regular hexagon.
Exercise:
Discriminant = -4 with two distinct complex roots 2i and -2i
The two roots of the equation are -1 + i and -1 -i.
We have to find the two square roots of 3 + 4i.
The modulus is 5; the argument is 0.9273
The square roots have modulus 2.2360 and argument 0.463647 or -0.463647.
The two roots of the equation are 2+i and -2-i.
Discriminant = 4.
The two roots of the equation are i and i-2.
Find sum and product of the roots of ix2 +(2-i)x + 3 -2i = 0
The sum is -(2-i)/i = 1 + 2i
The product is (3 -2i)/i = -2 -3i
Divide x3- i x2 + (1+i)x +1 by (x-i)
| 1 -i 1+i 1
i | i 0 i-1
--------------------------
1 0 1+i i
The quotient is x2 +1+i and the remainder is i
x3 - i x2 + (1+i)x +1 | x - i
x3 - i x2 |________________
----------------------- x2 + (1+i)
(1+i)x +1
(1+i)x -i + 1
--------------
i
The quotient is x2 +1+i and the remainder is i
| 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.
Remark: One can prove that the set of the roots of a polynomial equation does not depend
on the order in which we find the 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.
Exercise
Solution:
If the quadratic equation has real coefficients, then the conjugate complex number 2+3i is the other root.
The sum of these roots is 4 and the product is 13. The required quadratic equation is x2 -4x + 13 = 0.
Say P(x) = a x5 + bx4 + cx3 + dx2 + ex + fExample:
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 g h i j k) Hence , -a(g+h+i+j+k) = b and a((-1)5 g h i j k) = 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 a xn + b xn-1 + ... + l of degree n, we have The product of the roots is (-1)n l/a .
