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.
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
i.i = -1 or i^{2} = -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(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
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) = ----------------------- (c^{2} + d^{2}) (ac + bd) (bc - ad) = --------- + i----------- (c^{2} + d^{2}) (c^{2} + d^{2}) |
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 = c^{2} = (-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 -a^{2}.
a.b.i and -a.b.i are the square roots of - a^{2}b^{2}.
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) <=> x^{2} - y^{2} + 2xyi = a + bi (2) <=> x^{2} - y^{2} = a and 2xy = b (3) Because b is not 0, y is not 0 and so <=> x^{2} - y^{2} = a and x = b/(2y) b^{2} b <=> ---- - y^{2} = a and x = ---- (4) 4y^{2} 2yThe first equation of (4) gives us y and the second gives the corresponding x-value. Let t = y^{2} in the first equation of (4) then
4t^{2} + 4at - b^{2} = 0 (5) Let r = modulus of a + bi The discriminant = 16(a ^{2}+ b^{2}) = 16r^{2} We note the roots as t_{1} and t_{2}. <=> t_{1} = (- a + r)/2 and t_{2} = (- a - r)/2 (6)Since y is real and r > a, t_{1} > 0 and gives us values of y.
y_{1} = sqrt((r - a)/2) and y_{2} = -sqrt((r - a)/2) (7)The corresponding x values are
x_{1} = b/(2.y_{1}) and x_{2} = b/(2.y_{2}) (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 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(a^{2} + b^{2}) 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.
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 c_{1}=1+i and c_{2}=2-i in the Gauss-plane.
Then draw the product c_{3}=3+i.
Measure the modulus of each of those numbers and see that |c_{1}|.|c_{2}| = |c_{3}|.
Measure the argument of each of those numbers and see that
argument of c_{1} + argument of c_{2} = argument of c_{3}.
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)) r^{2} rRule: 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))
( r (cos(t) + i sin(t)) )^{n} = r^{n} .(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)
(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.
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 ix^{2} +(2-i)x + 3 -2i = 0
The sum is -(2-i)/i = 1 + 2i
The product is (3 -2i)/i = -2 -3i
Divide x^{3}- i x^{2} + (1+i)x +1 by (x-i)
| 1 -i 1+i 1 i | i 0 i-1 -------------------------- 1 0 1+i iThe quotient is x^{2} +1+i and the remainder is i
x^{3} - i x^{2} + (1+i)x +1 | x - i x^{3} - i x^{2} |________________ ----------------------- x^{2} + (1+i) (1+i)x +1 (1+i)x -i + 1 -------------- i The quotient is x^{2} +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 x^{3} + b x^{2} + d x + e Since c is a root of P(x) = 0 , we have a c^{3} + b c^{2} + d c + e = 0 => conj(a c^{3} + b c^{2} + 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 x^{2} -4x + 13 = 0.
Say P(x) = a x^{5} + bx^{4} + cx^{3} + dx^{2} + ex + fExample:
We factor this polynomial: P(x) = a(x-g)(x-h)(x-i)(x-j)(x-k) Then P(x) = a(x^{5} - (g+h+i+j+k)x^{4} + ...+(-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 x^{n} + b x^{n-1} + ... + l of degree n, we have The product of the roots is (-1)^{n} l/a .