International Journal of Research in Advent Technology (IJRAT)

1. International Journal of Research in Advent Technology, Vol.7, No.1, January 2019
E-ISSN: 2321-9637
Available online at www.ijrat.org
29
Analysis of complex roots of the equation
0)1(2)1(3  kxkxkx through complex continued
fraction, where 1k
Krithika S.1
and Gnanam A.2
Assistant professor1
, Assistant professor2
1
Department of Mathematics, Seethalakshmi Ramaswami College, Trichy-2.
2
Department of Mathematics, Government Arts College, Trichy-22.
Email:krithikasrc@gmail.com1
,gnaanam@yahoo.com2
Abstract
Complex roots of a polynomial always occur in pairs. To solve a cubic equation using Newton-Raphson method is
more advantageous than any other method. Here in this paper an attempt has been made to find the complex roots
of the equation 0)1(
2
)1(
3
 kxkxkx through complex continued fraction also a comparison is made
regarding number of iterations when 1k .
Keywords: Continued fraction, Finite complex continued fraction, Infinite complex continued fraction, Periodic
complex continued fraction.
Subject Classification: MSC 11A05, 11A55, 11J70, 11Y65, 30B70, 40A15.
1.INTRODUCTION
The Indian mathematician Aryabhata used a
continued fraction to solve a linear indeterminate
equation. For more than a thousand years, any work
that used continued fractions was restricted to
specific examples. Throughout Greek and Arab
mathematical writing, we can find examples and
traces of continued fractions. Euler showed that
every rational can be expressed as a terminating
simple continued fraction. He also provided an
expression for e in continued fraction form. He used
this expression to show that e and 2
e are irrational
[4].
Continued fraction plays an important role in number
theory. It is used to represent the rational numbers to
an another form by using Euclidean algorithm.
An expression of the form

3
3
2
2
1
1
0
0
b
a
b
a
b
a
b
a
q
p




where ii ba , are real or complex numbers is called a
continued fraction [6].
An expression of the form

2. International Journal of Research in Advent Technology, Vol.7, No.1, January 2019
E-ISSN: 2321-9637
Available online at www.ijrat.org
30








na
ne
e
a
e
a
e
a
e
a
3
3
2
2
1
1
0
0
where ,,,, 3210 aaaa are in )(iZ and sei ' are units
of complex numbers,   ,...3,2,1,,,1,1  kiiek is
known as a complex continued fraction [5].
The complex continued fraction is commonly
expressed as 


3
3
2
2
1
1
0
a
e
a
e
a
e
a .
The quantities ,,,,
3
3
2
2
1
1
0
a
e
a
e
a
e
a are called the
elements of the complex continued fraction.
In a finite complex continued fraction the number of
elements are finite, where as an infinite continued
fraction have infinite number of quantities.
Therefore
n
n
a
e
a
e
a
e
a
e
a 


3
3
2
2
1
1
0 is known as
finite complex continued fraction and an expression





1
1
3
3
2
2
1
1
0
n
n
n
n
a
e
a
e
a
e
a
e
a
e
a is known as
an infinite complex continued fraction.
In a finite or infinite complex continued fraction
,,,, 3210 aaaa are called the partial quotients and
,,, 321 eee are known as the partial numerators.
The length of a finite complex continued fraction
n
n
a
e
a
e
a
e
a
e
a 


3
3
2
2
1
1
0 is 1n and the length
of a infinite continued fraction





1
1
3
3
2
2
1
1
0
n
n
n
n
a
e
a
e
a
e
a
e
a
e
a is  .
The value of the finite complex continued fraction is
denoted as 





n
n
a
e
a
e
a
e
a
e
a ,,,,
3
3
2
2
1
1
0 .
2. PROPERTIES AND ALGORITHM OF
COMPLEX CONTINUED FRACTION:
2.1 Convergence of complex continued fraction:[1]
The successive convergence of the complex
continued fractions are  0a , 





1
1
0 ,
a
e
a ,






2
2
1
1
0 ,,
a
e
a
e
a and so on and they are denoted by
0
0
0
q
p
c  ,
1
1
1
q
p
c  ,
2
2
2
q
p
c  ,… In general the nth
convergent is denoted by







n
n
n
n
n
a
e
a
e
a
e
a
e
a
q
p
c ,,,,
3
3
2
2
1
1
0 , where spi '
and sqi ' are called the numerators and denominators
of convergent of the complex continued fraction.
2.2 Properties of complex continued fraction:[1]
Let 





 ,,,,,
3
3
2
2
1
1
0
n
n
a
e
a
e
a
e
a
e
a be an infinite
complex continued fraction. We inductively define
two infinite sequences kp and kq , 1k by
00 ap  11 p
10 q 01 q
1,21   kpepap kkkkk
1,21   kqeqaq kkkkk
2.3 Algorithm of complex continued fraction:[2,5,7]
Let Cx . Suppose we wish to find continued
fraction expansion of x . Take )Im()Re(0 xixx  .
Let  )Re()Re( xxa  and  )Im()Im( xxb  so
that 0a and 0b .
Now        )Im()Re(0 xixx .

3. International Journal of Research in Advent Technology, Vol.7, No.1, January 2019
E-ISSN: 2321-9637
Available online at www.ijrat.org
31
The floor function  is defined as









bawithbifai
bawithbaif
baif
1
11
00

Set  00 xa  . Then  
 00
0
1
xx
x

 and
       )Im()Re( 111 xixx .
Again set  11 xa  . Then  
 11
2
1
xx
x

 and
       )Im()Re( 222 xixx .
 22 xa  . Then  
 22
3
1
xx
x

 and
       )Im()Re( 333 xixx ….
 11   kk xa .
Then  
 11
1
 

kk
k
xx
x and
       )Im()Re( kkk xixx .
and  kk xa  .
The algorithm terminates if the complex continued
fraction is finite.
3. FINDING COMPLEX ROOTS OF THE
EQUATION 0)1()1( 23
 kxkxkx
USING COMPLEX CONTINUED FRACTION:
Consider the equation
0)1()1( 23
 kxkxkx .It can be factored as
0)1()( 2
 xxkx
In [5] the complex roots of the quadratic equation
012
 xx have been found. The roots are given
in terms of complex continued fraction as
 iii  2,21,2,1 and
 iii 21,2,2 
Clearly the other root of the equation is kx  .
Taking 1k we now find complex roots of the
equation using continued fraction method and
Newton-Raphson method and compare them.
4. ILLUSTRATION:
To find the complex roots of the cubic equation
013
z :
Method I: Using Newton-Raphson method[3,8]
Let 01)( 3
 zzf
Taking iyxz  we get
0)3()13()( 3223
 yyxixyxiyxf
Consider
0)3(),(0)13(),( 3223
 yyxyxvandxyxyxu
. Then
and











)(36
6)(31
22
22
1
yxxy
xyyx
D
J where
22222222
)(936)(9 yxyxyxJD 
Using the Newton- Raphson method
kk
kk
FJxx 1)()1( 
 , we obtain


































)3(
)13(
)(36
6)(31
32
23
22
22
1
1
kkk
kkk
kkkk
kkkk
kk
k
k
k
yyx
yxx
yxyx
yxyx
Dy
x
y
x
Taking initial approximation as
)25.0,25.0(),( 00 yx































8333333.2
1666667.0
03125.0
96875.0
0375.0
375.00
140625.0
1
25.0
25.0
1
1
y
x
The successive iterations are given below
k ),( kk yx
0 (0.25, 0.25)
1 (0.1666667, 2.8333333)
2 (0.15220505, 1.89374026)
3 (0.19263553, 1.27724322)
















 22
22
336
633
yxxy
xyyx
vv
uu
J
yx
yx

4. International Journal of Research in Advent Technology, Vol.7, No.1, January 2019
E-ISSN: 2321-9637
Available online at www.ijrat.org
32
4 (0.31932197, 0.91041889)
5 (0.49252896, 0.83063199)
6 (0.49983161, 0.86738607)
7 (0.49999870, 0.86602675)
Therefore the approximation of the root is
0.5+i0.8660. The approximation of the second root is
0.5-i0.8660.
Method II: Using Continued fraction method
The number of roots of the cubic equation 013
z
is three in which one real root -1. Then the cubic
equation is reduced to the quadratic equation
012
 zz and it has a pair of complex roots.
In [5] we found that the complex continued fraction
of the roots of the given equation 012
 zz are
 iiiz  2,21,2,1 and
 iiiz 21,2,2  .
Using the properties of complex continued fraction
00 ap  11 p
10 q 01 q
1,21   kpepap kkkkk
1,21   kqeqaq kkkkk
we get
ip 0 and 0
0
0
00 1 a
q
p
cq 
ip 211  and i
i
q
p
cq 

 5.0
2
21
2
1
1
11
ip 332  and
i
i
i
q
p
ciq 8824.05.0
41
33
41
2
2
22 



ip 743  and
i
i
i
q
p
ciq 8615.05.0
74
74
74
3
3
33 



ip 4154  and
i
i
i
q
p
ciq 8636.05.0
1111
415
1111
4
4
44 



305 p and
i
iq
p
ciq 8660.05.0
2615
30
2615
5
5
55 



Hence one complex root is i8660.05.0  .
Obviously the pair of the above root is i8660.05.0  .
From the above two methods it is observed that the
number of iterations required in complex continued
fraction method is fewer than the Newton- Raphson
method.
5. CONCLUSION
The complex roots of the equation
0)1()1( 23
 kxkxkx for 1k can be
solved using Newton-Raphson method and complex
continued fraction method. Factorization of this
equation leads to the factors kx  and
012
 xx . In [5] complex solutions of
012
 xx in terms of complex continued
fraction were calculated. As a special case taking
1k and comparing the solutions, we observe that
the number of iterations required in complex
continued fraction method is fewer than the Newton-
Raphson method.
REFERENCES
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Continued Fraction Algorithms”, Thesis,
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[2] Ivan Niven , Herbert S. Zuckerman, Hugh L.
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International Publishers, New Delhi.
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Shallit, Wadim Zudilin, “Neverending
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fractions”, Cambridge University Press.
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integer and 






2
n
k ”, International Journal of
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[6] Neville Robbins, “Beginning Number Theory”,
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