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
Standard Form of Complex Numbers
a + ib
Complex numbers are made up of a real number part and an imaginary number part.
In this form, a is the real number part and b is the imaginary number part.
Note that either one of these parts can be 0.
An example of a complex number written in standard form is – 10 + 5i
Equality of Complex Numbers
a + ib = c + id
if and only if a = c AND b = d.
In other words, two complex numbers are equal to each other if their real numbers match AND their imaginary numbers match.
Addition and Subtraction of Complex Numbers
(a + ib) + (c + id) = (a + c) + (b + d)i
(a + ib) - (c + id) = (a - c) + (b - d)i
In other words, when you add or subtract two complex numbers together, you add or subtract the real number parts together, then add or subtract their imaginary parts together and write it as a complex number in standard form.
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.
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.
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.
Since the product of the roots of (5) is (-b2/4) < 0 , t2 is strictly negative.
So we find two values of y. We note these values y1 and y2.
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 :
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))
----------------------- = 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.
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)
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.
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.
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.
