An introduction to complex numbers




The complex numbers

Are the real numbers not sufficient?

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 complex number

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 representation of a complex number

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.

Equal complex numbers

Two complex numbers (a,b) and (c,d) are equal if and only if (a = c and b = d).
So

a + bi = c + di
<=>
a = c and b = d


Sum of complex numbers

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.

 
                

Product of complex numbers

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

A special product of complex numbers

(0,1).(0,1)=(-1,0) or the equivalent

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!

Notation, sum and product

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

Opposite complex numbers

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

Subtraction

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

Conjugate complex numbers

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

Modulus of a complex number

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


The structure C, + , .

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.

Multiplication in practice

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

Examples

(2+3i). (5 - 2 i) = 16 + 11i
i.(3-i).i = -3 + i

Division in practice

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)



Examples

 
16+11i
------ = 2 + 3i
5 - 2i

15-7i
----- = -7 - 15 i
 i

Square roots and complex numbers

Square roots of a positive real number

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.

Square roots of strict negative real number

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.

Square roots of a complex number a + ib that is not real

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.
 
        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)
Here, r is the modulus of a + bi.

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 - 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)

Polar representation of complex numbers

modulus and argument of a complex number

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.

 
               

Polar form of complex numbers

We just saw that s( cos(t) , sin(t) )
and we have the vector-equation OP = r. OS
Therefore P( r cos(t) , r sin(t) ) but also P(a,b).
It follows that a = r cos(t) ; b = r sin(t).
So we have a + ib = r cos(t) + i r sin(t) or

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) )

Equal numbers in polar representation

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

Product in polar representation

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.

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'))


With this rule we have a geometric interpretation of the multiplication of complex numbers.

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.

Conjugate numbers in polar representation

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))

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.

The inverse of a complex number in polar representation

We have :
 
        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.

Quotient two complex number in polar representation

We have :
 

 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))

Natural power of a complex number

Using the multiplication rule we have:

( 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)

Formula 'De Moivre'

Take the previous formula with r = 1.

(cos(t) + i sin(t))n = cos(nt) + i sin(nt)


Negative power of a complex number

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))
Example : (cos(0.1) + i sin(0.1))-10 = (cos(-1) + i sin(-1)) = (cos(1) - i sin(1))

Other properties of complex numbers

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'|

n-th root of a complex number

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 + 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.
These n-th roots are : r1/n( cos(t/n + 2kpi/n) + i sin(t/n + 2kpi/n) ) with k = 0, ... ,n-1
and with r1/n > 0.

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:
Draw, in the Gauss-plane, a random complex number with modulus 4.
Construct the two square roots of this complex number.

Polynomials and complex numbers

Properties

It can be shown that many formulas and properties for polynomials with real coefficients also hold for polynomials with complex coefficients. Examples: Examples:
  1. Solve : x2 + 2 x + 2 = 0

    Discriminant = -4 with two distinct complex roots 2i and -2i
    The two roots of the equation are -1 + i and -1 -i.

  2. Solve : x2 - 4 i - 3 = 0

    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.

  3. Solve : x2 + (2 - 2i) x - 1 - 2i = 0

    Discriminant = 4.
    The two roots of the equation are i and i-2.

  4. 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)

  5. Find c such that x3- c x2 + (1+i)x +1 is divisible by x+2i
    We substitute x by (-2i) in x3- c x2 + (1+i)x +1.
    We find 4 c + 6 i + 3. This remainder must be 0.
    c = -3/2 i - 3/4

Theorem of d'Alembert

Each polynomial equation with complex coefficients and with a degree n > 0, has at least one root in C.

Number of roots of a polynomial 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.
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.

Roots of a polynomial equation with real coefficients.

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.
Exercise
Find a quadratic equation with real coefficients, such that 2 -3i is a root.

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.

Factorization of a polynomial with real coefficients

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.

Sum and product of the roots of a polynomial equation

 
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 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 .
Example:
Sum and product of the roots of x4 + i x3 -(1-i)x2 +3x - 4 is -i and -4

 

Solved problems about complex numbers

 
You can find solved problems about complex numbers using this link
 





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