Thursday, December 14

Complex Number Roots of Polynomials

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All polynomial equations have real and complex number solutions. In addition, positive  real number has two square roots as real number solutions. As well, all positive and negative real numbers have  nth root. They may have real diiferent n real and complex roots. In addition, complex numbers also have square and nth roots like real numbers. However complex nth roots are complex numbers.

Say one wants to solve the equation x cube -1 = 0 . In this instance, x = cube root of 1. That is one solution to the cube root of 1 is 1 itself. However, there are 2 other cube roots. They are complex numbers. A complex number have two portions. One is a real number part and other is an imaginary part. That is a complex number can be represnted by  ( x + iy), where i = square root of (-1). In addition, real numbers can be expressed as complex numbers.

1 = ( 1 + i*0). Say the cube root of 1 is ( x + iy). Then it can be expressed as follows:

( 1+ i*0) = ( x + iy)* (x+ iy) * ( x + i y) = ( x cube – 3 xy squared) + i ( 3 x squared y – y cube)

That is 3 x squared – y cube = 0 and  x cube – 3 x y squared = 1

That is substituting x = = or – y/square root of 3 in the equation x cube – 3 xysquared = 1 This equation has a real number solution of – square rootof 3/2. In addition, thre fore x = -1/2. That is one of the complex number cube root of 1 is (-1/2 – squareof 3/2 i).

In the same way, substituting x = -y/ square root of 3 , y= square root of 3/2. However x = -1/2. There fore the other complex number is -1/2 + square root of 3/2.

That in other words, the polynomial equation x cube -1 = 0 has three solution. One is a real number and other two are complex numbers.

In the same way, one can get the nth root of any number. However, as the n is getting larger this method becomes time consuming.

Polar representation of complex numbers

Say a complex number is 4 + 3 i , it can be reprented by a polar co-ordinate system. That is, in trigonometric form as below:

4 + 3 i = square root of 25* ( 4/ square root of 25 + i * 3/ square root of 25) . In this square root of 25 is called a modulus of the complex number. If cos (a) = 4/ square root of 25, in a rectangle triangle and the base is 4 units and perpendicular side is 3 units Cos ( a) = 4/ square root of 25. Then the sin (a) = 3/ square root of 25). That is any complex number can be represented  as follows:

x + iy = r ( cos (a) + i sin (a)) where r is a modulus of the complex number and (a) is called an argumnet of the complex number.  r= square root of x squared + y squared.

If  x squared + 1 = 0 .  x squared = -1, then  two square roots are complex numbers. However, as defined above square of -1 is (i) there fore the the solution is ( 0 + 1i) and ( 0 – 1i). That is, (i) and ( -i) are the roots of (-1). In general all negative real numbers only has complex square roots.

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