Friday, 30 May 2014

Why Complex Numbers?

ComplexNos

Complex numbers are introduced in high school mathematics today. A common question that springs in our mind is:

Why do we need complex numbers? We all know x2 > 0 for all non-zero real numbers. Then why bother to seek a ‘‘number’’ x such that x2 = -1?

First note that it is very clear that there is no real number satisfying x2 = −1. Till some point in the history when people encountered such an equation, they ignored it as being absurd. It was Gerolamo Cardano (1545) who pursued √−1 as an “imaginary” number and Rafael Bombelli (1572) developed it as a number system. This blog is an attempt to recall the motivation behind their interest in a “number” that was not “real”.

The “imaginary” numbers were introduced as a mathematical tool to make life simple in the real world. In contrast to quadratic equations, it is impossible to compute real roots of certain cubic equations without solving for x2 =−1. Thus, it was observed that even to obtain real roots of certain cubic equations one has to go outside the realm of real numbers.

It is known for, at least, 2300 years that the formula to compute roots of the quadratic equation ax2 + bx + c =0, for given a,b,c∈ℝ with a ≠ 0 is

x  =  
 − b ± 
b2 −4ac
2a
.

A straight forward way of deriving this formula

a x2 + bx + c=0
x2 + 
b
a
x
=
− 
c
a
x2 + 
b
a
x + 


b
2a



2



 
=
c
a
 + 


b
2a



2



 



x + 
b
2a



2



 
=
b2 − 4ac
4a2
x + 
b
2a
=
± 
b2 − 4ac
2a
x=
 − b ± 
b2 −4ac
2a
.

The case b2 −4ac < 0 always corresponds to non-existence of roots. Geometrically, in this case, the graph of the quadratic function never intersects x-axis. This situation sits well with our understanding that it is absurd to consider square root of negative numbers.

Let us derive the formula for roots of quadratic equation in an alternate way. This alternate approach will help us in deriving a formula for cubic equation. Note that if b=0 in the quadratic equation, then a x2 + c=0 and

x = 
c
a
.

Let us seek a ξ such that replacing x in ax2 + bx+c with y − ξ changes the equation to A y2 + C. Set x= y −ξ. Then

a x2 + bx + c=a (y−ξ)2 + b (y −ξ) + c
 =a y2 − 2aξ y + a ξ2 + by − b ξ +c
 =a y2 + (b − 2aξ)y + a ξ2 − bξ +c.

We shall choose ξ such that the coefficient of y is zero. Therefore, we choose ξ = b/2a. Thus,

a y2 + 
b2
4a
 − 
b2
2a
 + c = a y2 −
b2
4a
 + c.

The roots of a y2b2/4a + c =0 are

y = ± 
b2 − 4ac
2a

and the roots of ax2 + bx +c =0 are

x = − 
b
2a
± 
b2 − 4ac
2a
.

Let us employ the above approach to find a formula for roots of the cubic equation ax3 + bx2 +cx +d =0. We seek for a ξ such that replacing x in a x3 + b x2 +cx +d with y −ξ changes the equation to Ax3 + Cx +D. Then

a x3 + bx2 + cx + d=a (y−ξ)3 + b (y −ξ)2 + c (y− ξ) + d
 =a y3 − 3aξ y2 + 3 a ξ2 y − ξ3 + by2 − 2b ξ y +b ξ2
   + cy − c ξ +d
 =a y3 + (b − 3aξ)y2 + (3 a ξ2 − 2 bξ +cy +bξ2
   − ξ3 − c ξ +d.

Demanding ξ to be such that the coefficients of both y2 and y vanish is too restrictive because in that case we must have b − 3aξ = 3a ξ2 − 2b ξ +c =0. This would imply that ξ = b/(3a) and b2 = 3ac which is too restrictive for a general cubic equation.

Let us eliminate the coefficient of y2, by choosing ξ = b/(3a). Then the cubic equation in y has the form

a y3 + 


a c − b2
3a



y + 


b
3a



3



 
(3a −1) + 
3ad − bc
3a
 = 0.

Now, set

p := 
a c − b2
3a

and

q := 


b
3a



3



 
(3a −1) + 
3ad − bc
3a

to make the equation appear as a y3 + py +q =0. To reduce this cubic equation to a quadratic equation in a new variable we use the Vieta’s substitution which says define a new variable z such that

y = z − 
p
3az
.

The Vieta’s substituion can be motivated but we shall not digress in to that domain. Then the equation a y3 + py +q =0 transforms to a quadratic equation

27 a4 (z3)2 + 27 q a3 z3 − p3=0.

The roots of this equation are

z3 = 
 − q ± 
q2 + 
4p3
27 a2
2a
.

At first glance, it seems like z has six solutions, but three of them will coincide. Thus, finding z leads to finding y and, hence, to x. Thus far, we have only derived a formula for roots of cubic equation. We are yet to address the question of complex numbers.

As a simple example, consider the cubic function x3 − 3x. The roots of this equation can be easily computed by rewriting x3x = x(x2 −3) = x (x +√3)(x−√3) and hence has exactly three roots 0, √3, −√3.

Let us try to compute the roots of x3x using the formula derived above. Note that x3x is already in the form with no x2 term. Thus, a=1, p=−3, q=0 and z3 = ± √−1. The cubic equation has real roots but z is not in the realm of real numbers. √−1 makes no sense, since square of any non-zero real number is positive. This is the motivation for complex numbers! We set i:= √−1 and, hence, i2 = −1. We introduce this notation to avoid using the property of square root, √ab = √ab. Negative numbers will not inherit this property because √−1−1 = −1 ≠ 1 = √(−1)(−1). With this setting, z3 = ± i. Now recall from your course on complex numbers that the cube roots of i are z = −i, √3+i/2, −√3+i/2. Using this in the formula

x = z + 
1
z

we get x = 0, √3, − √3. The cube roots of −i are z = i, √3i/2, −√3i/2 which yield the same roots. Expanding to complex number system helped us in solving a real cubic equation with only real roots! This little bold step helped us understand/expand in many ways. Complex numbers and complex functions has found application in engineering and other sciences, especially, via analytic functions and analytic continuations.