Communications on Applied Nonlinear Analysis 

ISSN: 1074-133X 

Vol 31 No. 7s (2024) 

 

466 https://internationalpubls.com 

Common Fixed Point Results for Contractive Mappings in Bicomplex 

Valued B-Metric Spaces 

 

Md. Azizul Hoque
 

Department of Mathematics, Sreegopal Banerjee College, 

Mogra, Dist-Hooghly , PIN- 712148, West Bengal, India. Email:  mhoque3@gmail.com 

 

Article History: 

Received: 01-06-2024 

Revised: 03-07-2024 

Accepted: 29-07-2024 

Abstract:  

In this article, we extend and generalised the results of Ahmad et.al., and to establish the  

existence and uniqueness of common fixed points for pair of self mappings on a closed ball 

in bicomplex valued b-metric space. Our results generalised well known results in the 

literature.  

Keywords: Common fixed point, bicomplex valued metric space. 

2010 MSC:47H09;47H10;30G35;46N9;54H25. 

 

1. Introduction 

The theory of bicomplex numbers have been studied for quite a long time, which probably began 

with the works [3,4,5].The algebra of bicomplex numbers are widely used in the literature as it 

becomes a viable commutative alternative [4,5] to the non commutative skew field of quaternions 

(both are four-dimensional and generalization of complex numbers).The commutativity in the former 

is gained at the cost of the fact that the ring of these numbers contains zero-divisors and so can not 

form a field .It is well known that the fixed point theory plays a very important role in theory and 

applications, in particular, whose importance comes from finding roots of algebraic equation and 

numerical analysis. Banach contraction principle in [15] gives appropriate and simple conditions to 

establish the existence and uniqueness of a solution of an operator equation 𝑇�𝑥� = 𝑥�. Later, a number 

of papers were devoted to the improvement and generalization of that result. Most of these results 

deal with the generalizations of the different contractive conditions in metric spaces [7,10,11,17]. 

There have been a number of generalizations of metric spaces such as vector valued metric spaces, 

𝐺�- metric spaces, pseudometric spaces, fuzzy metric spaces, 𝐷�-metric spaces, cone metric spaces, 

and modular metric spaces. Bakhtin [14] introduced the notion of 𝑏�-metric space which is a 

generalized form of metric spaces. Azam et al. [2,9] introduced the notion of complex-valued metric 

space which is a generalization of classical metric space and established sufficient conditions for the 

existence of common fixed points of a pair of mappings satisfying a contractive condition. The 

concept of complex valued 𝑏�-metric spaces was introduced in 2013 by Rao et al. [18]. In sequel, 

Mukheimer [16] proved some common fixed point theorems in complex valued 𝑏�-metric spaces. 

Recently  Junesang Choi et al.  [ 1] introduced the notion of  bi-complex valued metric space which 

is a generalization of classical metric space and proved certain common fixed point theorems for a 

pair of weakly compatible mappings satisfying (CLRg) (or (E.A)) property in the bicomplex valued 

metric spaces. In 2019 Jebril et.al.[8] proved some important theorems on common fixed point 

theorems under rational contractions for pair of mappings in bicomplex valued metric spaces . In this 



Communications on Applied Nonlinear Analysis 

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article , we extend and generalised the results of Ahmad et.al.[13], Dubey et al. [17] and Rao et al. 

[18] and to establish the  existence and uniqueness of common fixed points for pair of self mappings 

on a closed ball in bicomplex valued b-metric space which extends a recent results of, I.Beg, 

S.K.Datta and D.Pal [6] , Md. A.Hoque [12] and several others.Here in the following, the set of 

bicomplex numbers ,complex numbers and real numbers are denoted by ℂ2, ℂ and ℂ0 respectively. 

ℂ2 becomes a real commutative algebra with the identity  

1=1+𝑖1. 0+𝑖2.0+𝑖1𝑖2�.0, the set of bicomplex number is defined as  

ℂ2={ξ=𝑎0+𝑎1𝑖1+𝑖2.�𝑎2+𝑖1𝑖2�.�𝑎3:�𝑎0,𝑎1,�𝑎2, 𝑎3 ∈ ℂ0𝑎𝑛𝑑�𝑖1
2 = 𝑖2

2 = −1}. 

Definition 1:Let  ξ1=𝑢1+𝑖2�𝑢2 ∈ ℂ2 and ξ2=𝑣1+𝑖2�𝑣2 ∈ ℂ2 define partial order relation  ≲𝑖2 ��on ℂ2 as 

follows (see, e.g. [6]): �ξ1 ≲𝑖2 � ξ2   if and only if �𝑢1 ≲�𝑣1� �𝑎𝑛𝑑���𝑢2 ≲�𝑣2              ………..(1) 

where  ≲� is the partial order on ℂ1 (see,e.g [2]). 

Thus  ξ1 ≲𝑖2 � ξ2 if any one of the following properties holds: 

[b𝑜1] if �𝑢1 =�𝑣1� �𝑎𝑛𝑑���𝑢2 = 𝑣2; 

[b𝑜2] if �𝑢1 ≺�𝑣1� �𝑎𝑛𝑑���𝑢2 = �𝑣2; 

[b𝑜3] if �𝑢1 =�𝑣1� �𝑎𝑛𝑑���𝑢2 ≺ �𝑣2; 

[b𝑜4] if  𝑢1 ≺�𝑣1� �𝑎𝑛𝑑���𝑢2 ≺�𝑣2. 

We write ξ1 ≰𝑖2 � ξ2 if ξ1 ≲𝑖2 � ξ2 and ξ1 ≠� ξ2 i.e , one of [b𝑜2], [b𝑜3] and [b𝑜4] is satisfied and we 

write  ξ1 ≺𝑖2 � ξ2 if only   [b𝑜4] is satisfied. 

The norm ||.||: ℂ2 → ℂ0
+

  (the set of all non negative real numbers) of a bicomplex number is defined 

as || ξ||=ℂ1√𝑎0
2 + 𝑎1

2 + 𝑎2
2 + 𝑎3

2 …………………(2) 

For any two bicomplex numbers  ξ1 , ξ2 ∈ ℂ2, one can easily verify that  

0 ≲𝑖2 ξ1 ≲𝑖2 � ξ2 ⇒ ||ξ1|| ≤ ||ξ2||; ||ξ1 +�ξ2|| ≤ ||ξ1|| + ||ξ2||; ||ξ1. ξ2|| ≤ √2||ξ1||. ||ξ2|| and  

 ||a ξ||≤�a||�ξ|| where a∈ ℂ0
+

.  

Bicomplex metric space: 

Choi et al. [1] define the bicomplex valued metric space as : 

Definition 2:Let X be a non empty set. Suppose the mapping d: X ×X → ℂ2 satisfies the following 

conditions: 

[1] 0 ≲𝑖2 d(x,y) for all x,y∈ 𝑋; 

[2] d(x,y)=0 if and only if x=y; 

[3] d(x,y)=d(y,x)  for all x,y∈ 𝑋; 

[4] 𝑑(𝑥, 𝑦) ≲𝑖2 d(x,z)+d(z,y) for all x,y∈ 𝑋 . 

Then (X,d) is called a bicomplex valued metric space. 



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Definition 3:[1]  A sequence in a nonempty set X is a function x: ℕ → ℂ2, which is expressed by its 

range set {𝑥𝑛} where x(n)=�𝑥𝑛�(n∈ ℕ ). Let {𝑥𝑛} be a sequence in bicomplex valued metric space 

(X,d). The sequence {𝑥𝑛} is said to converge to x∈ 𝑋if and only if for any 0 ≺𝑖2 ℰ ∈ ℂ2, there exists 

N∈ ℕ depending on ℰ such that d(𝑥𝑛,x)�≺𝑖2 ℰ as n>N.. A sequence {𝑥𝑛} in a bicomplex valued 

metric  space (X,d) is said to be Cauchy sequence if and only if  for any 0 ≺𝑖2 ℰ ∈ ℂ2 ,there exists N 

∈ ℕ depending on ℰ  such that d(𝑥𝑛,�𝑥𝑚)�≺𝑖2 ℰ as n,m>N. A bicomplex valued metric space is said to 

be complete if and only if  every Cauchy sequence in X converges in X. 

Definition 4.  Let 𝑋� be a nonempty set and let 𝑠�≥1 be a given real number. A function  

 d: 𝑋�×𝑋� → ℂ2  is called a bicomplex valued 𝑏�-metric on 𝑋� if for all 𝑥�, 𝑦�, 𝑧� ∈ 𝑋� the following 

conditions are satisfied: (i) 0 ≲𝑖2 (𝑥�, 𝑦�) and 𝑑�(𝑥�, 𝑦�) = 0 if and only if 𝑥�=𝑦�;  

(ii)d (𝑥�, 𝑦�) = d(𝑦�, 𝑥�); 

 (iii) d(𝑥�, 𝑦�) ≲𝑖2 s[d(𝑥�, 𝑧�) + d(𝑧�, 𝑦�)].  

The pair (𝑋�, 𝑑�) is called a complex valued 𝑏�-metric space.  

Example :  If 𝑋� = [0, 1], define the mapping  d: 𝑋�×𝑋�→ ℂ2 by  

d(𝑥�, 𝑦�) =(1 + i1 + i2 + i1i2)|x� − �y|2   , for all 𝑥�, 𝑦�  ∈ 𝑋�. Then (𝑋�, 𝑑�) is bicomplex valued 𝑏�-

metric space with 𝑠�=2. 

2. Main Results 

Theorem 1: Let (X,d) be a complete bicomplex valued metric space with coefficient s≥ 1 and 

𝑥0 ∈ 𝑋. 0< 𝑟 ∈ ℂ and A,B,C,D and E are non negative reals such that  

A+√2𝐵 + √2𝐶 + √2𝑠𝐷 + √2𝑠𝐸 < 1. Let S,T:X→X are mapping satisfying  

d(Sx,Ty) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐶

𝑑(𝑦,𝑆𝑥)𝑑(𝑥,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐷

𝑑(𝑥,𝑆𝑥)𝑑(𝑥,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐸

𝑑(𝑦,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

1+𝑑(𝑥,𝑦)
         (1.1) 

for all x,y∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ . If ‖𝑑(𝑥0, 𝑆𝑥0‖ ≲𝑖2 (1 − 𝜆)|𝑟| where λ= max{
𝐴+√2𝑠𝐷

1−𝐵−√2𝑠𝐷
,

𝐴+√2𝑠𝐸

1−√2𝐵−√2𝑠𝐸
+,         

(1.2) 

Then there exist a unique point u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  such that u=Su=Tu. 

Proof: Let 𝑥0�be an arbitrary point in X and define 𝑥2𝑛+1 = 𝑆𝑥2𝑛 and 𝑥2𝑛+2 = 𝑇𝑥2𝑛+1 where 

n=0,1,2….   We will prove that 𝑥𝑛 ∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  for all n∈ℕ by mathematical induction . Using 

inequality (1.2) and the fact that  λ= max{
𝐴+√2𝑠𝐷

1−𝐵−√2𝑠𝐷
,

𝐴+√2𝑠𝐸

1−√2𝐵−√2𝑠𝐸
+ < 1 we have 

‖𝑑(𝑥0, 𝑆𝑥0‖ ≲𝑖2
|𝑟|�.�It implies that 𝑥1 ∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ .  Let 𝑥2, 𝑥3, … , 𝑥𝑘 ∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ ��for some kєℕ. 

If k=2n+1 where n=0,1,2…,
𝑘−1

2
 or k=2n+2 where n=0,1,2,…., 

𝑘−2

2
,  we obtain by using inequality 

(1.1)  

d(x2n+1, x2n+2) = d(Sx2n, Tx2n+1) 

                               ≲𝑖2 𝐴�d(x2n, x2n+1) + 𝐵
d(x2n+1,Tx2n+1)d(x2n,Sx2n)

1+d(x2n,x2n+1)
+C

d(x2n+1,Sx2n)d(x2n,Tx2n+1)

1+d(x2n,x2n+1)
 



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                                       +D
d(x2n,Tx2n+1)d(x2n,Sx2n)

1+d(x2n,x2n+1)
+ 𝐸

d(x2n+1,Tx2n+1)d(x2n+1,Sx2n)

1+d(x2n,x2n+1)
 

                              ≲𝑖2 𝐴�d(x2n, x2n+1) + 𝐵
d(x2n+1,x2n+2)d(x2n,x2n+1)

1+d(x2n,x2n+1)
+D

d(x2n,x2n+2)d(x2n,x2n+1)

1+d(x2n,x2n+1)
 

This implies  

‖d(x2n+1, x2n+2)‖ ≤ 𝐴‖d(x2n, x2n+1)‖ 

                                                         

+√2𝐵
‖d(x2n+1,x2n+2)‖‖d(x2n,x2n+1)‖

‖1+d(x2n,x2n+1)‖
+ √2𝐷

‖d(x2n,x2n+2)‖‖d(x2n,x2n+1)‖

‖1+d(x2n,x2n+1)‖
 

Since ‖1 + d(x2n, x2n+1)‖ > ‖d(x2n, x2n+1)‖ 

Hence ‖d(x2n+1, x2n+2)‖ ≤ 𝐴‖d(x2n, x2n+1)‖ + √2𝐵‖d(x2n+1, x2n+2)‖ + √2𝐷‖d(x2n, x2n+2)‖ 

                                                ≤ 𝐴‖d(x2n, x2n+1)‖ + √2𝐵‖d(x2n+1, x2n+2)‖ 

                                                                                 +�√2𝑠𝐷*‖d(x2n, x2n+1)‖ + ‖d(x2n+1, x2n+2)‖    

(1-√2𝐵 −�√2𝑠𝐷)‖d(x2n+1, x2n+2)‖ ≤ (𝐴 + √2𝑠𝐷)�‖d(x2n, x2n+1)‖ 

⇒‖d(x2n+1, x2n+2)‖ ≤
𝐴+√2𝑠𝐷

1−√2𝐵−√2𝑠𝐷
�‖d(x2n, x2n+1)‖                                                               (1.3) 

Similarly we get, 

‖d(x2n+2, x2n+3)‖ ≤
𝐴+√2𝑠𝐸

1−√2𝐵−√2𝑠𝐸
�‖d(x2n+1, x2n+2)‖                                                               (1.4) 

Putting  λ= max{
A+√2sD

1−B−√2sD
,

A+√2sE

1−√2B−√2sE
+, we�obtain� 

‖d(xk, xk+1)‖ ≤ 𝜆𝑘‖𝑑(𝑥0, 𝑥1‖                                                                                                        (1.5) 

For all kєℕ 

‖d(x0, xk+1)‖ ≤ 𝑠‖𝑑(𝑥0, 𝑥1‖ + 𝑠‖d(x1, xk+1)‖ 

                          ≤ 𝑠‖𝑑(𝑥0, 𝑥1‖ + 𝑠2‖𝑑(𝑥1, 𝑥2‖ + 𝑠2‖𝑑(𝑥2, 𝑥𝑘+1‖      

                           ≤ 𝑠‖𝑑(𝑥0, 𝑥1‖ + 𝑠2‖𝑑(𝑥1, 𝑥2‖ + 𝑠3‖𝑑(𝑥2, 𝑥3‖ +⋯+ 𝑠𝑘+1‖𝑑(𝑥𝑘, 𝑥𝑘+1‖� 

                           ≤ 𝑠‖𝑑(𝑥0, 𝑥1‖+𝑠2𝜆‖𝑑(𝑥0, 𝑥1‖ + 𝑠3𝜆2 ‖𝑑(𝑥0, 𝑥1‖ +⋯+ 𝑠𝑘+1 𝜆𝑘 ‖𝑑(𝑥0, 𝑥1‖       

                          =‖𝑑(𝑥0, 𝑥1‖ [s+𝑠2𝜆 + 𝑠3𝜆2 +⋯+ 𝑠𝑘+1 𝜆𝑘 ]        

                            ≤ (1 − 𝜆)|𝑟|𝑠
1−(𝑠𝜆)𝑘+1

1−𝑠𝜆
              [as   ‖𝑑(𝑥0, 𝑥1‖ �≤ (1 − 𝜆)|𝑟|- 

                            ≤ |𝑟|  when s=1 

Gives   xk+1 ∈B( x0, 𝑟) . hence  𝑥𝑛 ∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅      ∀𝑛 ∈ ℕ  

and �‖d(xn, xn+1)‖ ≤ 𝜆𝑛‖𝑑(𝑥0, 𝑥1‖ ∀𝑛 ∈ ℕ                                                                             (1.6) 

without loss of generality , we take m> 𝑛 , then  



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   �‖d(xn, xm)‖ ≤ 𝑠‖d(xn, xn+1)‖ + 𝑠‖d(xn+1, xm)‖         

                           ≤ 𝑠‖d(xn, xn+1)‖ + 𝑠2‖𝑑(𝑥𝑛+1, 𝑥𝑛+2‖ + 𝑠2‖𝑑(𝑥𝑛+2, 𝑥𝑚‖  

                          ≤ 𝑠‖d(xn, xn+1)‖ + 𝑠2‖𝑑(𝑥𝑛+1, 𝑥𝑛+2‖ +⋯+ 𝑠𝑚−𝑛−1‖𝑑(𝑥𝑚−2, 𝑥𝑚−1‖ 

���������������������������������������������������������������������������������������������������������������������������+𝑠𝑚−𝑛‖𝑑(𝑥𝑚−1, 𝑥𝑚‖           

By using (1.6) we get, 

     �‖d(xn, xm)‖ ≤ 𝑠𝜆𝑛‖𝑑(𝑥0, 𝑥1‖ + 𝑠2𝜆𝑛+1‖𝑑(𝑥0, 𝑥1‖ 

������������������������������������+𝑠3𝜆𝑛+2‖𝑑(𝑥0, 𝑥1‖ +⋯+�𝑠𝑚−𝑛−1𝜆𝑚−2‖𝑑(𝑥0, 𝑥1‖ + 𝑠𝑚−𝑛𝜆𝑚−1‖𝑑(𝑥0, 𝑥1‖             

                             =∑ 𝑠𝑖𝜆𝑛+𝑖−1𝑚−𝑛
𝑖=1 ‖𝑑(𝑥0, 𝑥1‖              

                              ≤
𝑠𝜆𝑛

1−𝜆𝑠
 ‖𝑑(𝑥0, 𝑥1‖ →0 as m,n→∞     

This implies that the sequence {𝑥𝑛+�𝑎𝑠�a cauchy sequence in B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ . Therefore there exists a point 

u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  with lim𝑛→∞ 𝑥𝑛 = 𝑢. 

We prove that u=Su. Let us consider  

‖𝑑(𝑢, 𝑆𝑢)‖  ≤ 𝑠‖𝑑(𝑢, 𝑥2𝑛+2)‖ + 𝑠‖𝑑(𝑥2𝑛+2, 𝑆𝑢)‖   

                      ≤ 𝑠‖𝑑(𝑢, 𝑥2𝑛+2)‖ + 𝑠‖𝑑(𝑇𝑥2𝑛+1, 𝑆𝑢)‖            

                      ≤ 𝑠‖𝑑(𝑢, 𝑥2𝑛+2)‖ + 𝑠‖𝑑(𝑆𝑢, 𝑇𝑥2𝑛+1)‖     

                      ≤ 𝑠‖𝑑(𝑢, 𝑥2𝑛+2)‖ + 𝐴𝑠‖𝑑(𝑥2𝑛+1, 𝑢)‖ + 𝑠𝐵√2
‖𝑑(𝑥2𝑛+1,𝑇𝑥2𝑛+1)‖‖𝑑(𝑢,𝑆𝑢)‖

‖1+𝑑(𝑢,𝑥2𝑛+1)‖
  

�+𝑠𝐶√2
‖𝑑(𝑥2𝑛+1,𝑆𝑢)‖‖𝑑(𝑢,𝑇𝑥2𝑛+1)‖

‖1+𝑑(𝑥2𝑛+1),𝑢)‖
+ 𝑠𝐷√2

‖𝑑(𝑢,𝑆𝑢)‖‖𝑑(𝑢,𝑇𝑥2𝑛+1)‖��

‖1+𝑑(𝑢,𝑥2𝑛+1)‖
 + 𝑠𝐸√2

‖𝑑(𝑥2𝑛+1,𝑇𝑥2𝑛+1)‖‖𝑑(𝑥2𝑛+1,𝑆𝑢)‖

‖1+𝑑(𝑢,𝑥2𝑛+1)‖
         

Notice that , 

       lim𝑛→∞‖𝑑(𝑢, 𝑥2𝑛+2)‖ = lim𝑛→∞‖𝑑(𝑥2𝑛+1, 𝑢)‖ + lim
𝑛→∞

‖𝑑(𝑥2𝑛+1, 𝑆𝑢)‖ = 0              

Hence ‖𝑑(𝑢, 𝑆𝑢)‖  =0 that is u=Su  

Similarly , u=Tu 

For uniqueness assume that 𝑢∗ in B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅   is a another common fixed point of S and T. Then  

‖𝑑(𝑢, 𝑢∗)‖ ≤ ‖𝑑(𝑆𝑢, 𝑇𝑢∗)‖

≤ A‖𝑑(𝑢, 𝑢∗)‖ + 𝐵√2
‖𝑑(𝑢, 𝑆𝑢)‖‖𝑑(𝑢∗, 𝑇𝑢∗)‖

‖1 + 𝑑(𝑢, 𝑢∗)‖
+ 𝐶√2

‖𝑑(𝑢∗, 𝑆𝑢)‖‖𝑑(𝑢, 𝑇𝑢∗)‖

‖1 + 𝑑(𝑢, 𝑢∗)‖

+ 𝐷√2
‖𝑑(𝑢, 𝑆𝑢)‖‖𝑑(𝑢, 𝑇𝑢∗)‖

‖1 + 𝑑(𝑢, 𝑢∗)‖
+ 𝐸

‖𝑑(𝑢∗, 𝑆𝑢)‖‖𝑑(𝑢∗, 𝑇𝑢∗)‖

‖1 + 𝑑(𝑢, 𝑢∗)‖
 

                     ≤ 𝐴‖𝑑(𝑢, 𝑢∗)‖ + 𝐶√2‖𝑑(𝑢∗, 𝑢)‖           [as  ‖1 + 𝑑(𝑢, 𝑢∗)‖ > ‖𝑑(𝑢, 𝑢∗)‖ ] 

∴  ‖𝑑(𝑢, 𝑢∗)‖ ≤ (𝐴 + 𝐶√2)‖𝑑(𝑢, 𝑢∗)‖     



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This is a contradiction because (𝐴 + 𝐶√2) < 1. Hence 𝑢 = 𝑢∗. Therefore u is a unique common 

fixed point of T and S.  

This completes the proof of the theorem . 

Remark 1.1: The result of theorem 1.1 remains true if the condition (1.2) is replaced by the 

condition  ‖𝑑(𝑥0, 𝑇𝑥0‖ ≤ (1 − 𝜆)|𝑟|. 

Corollary  1.1: Let (X,d) be a complete bi-complex valued b-metric space with coefficient s≥ 1 and 

degenerated 1+d(x,y) , ‖1 + d(x, y)�‖ ≠ 0�and 𝑥0 ∈ 𝑋. Let 0  ≤ 𝑟 ∈ ℂ and A,B,C,D  are non negative 

reals such that A+√2𝐵 + √2𝐶 + √2𝑠𝐷 < 1. Let S,T:X→X  are mappings satisfying:  

d(Sx,Ty) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐶

𝑑(𝑦,𝑆𝑥)𝑑(𝑥,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐷

𝑑(𝑥,𝑆𝑥)𝑑(𝑥,𝑇𝑦)

1+𝑑(𝑥,𝑦)
���������������������������������������� 

for all x,y∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ . If ‖𝑑(𝑥0, 𝑆𝑥0‖ ≤ (1 − 𝜆)|𝑟| , where λ= max{
𝐴+√2𝑠𝐷

1−√2𝐵−√2𝑠𝐷
,

𝐴

1−√2𝐵
+,                 

then there exist a unique point u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  such that u=Su=Tu. 

Proof: We can prove this result by applying theorem 1.1 by setting E=0. 

Corollary  1.2: Let (X,d) be a complete bi-complex valued b-metric space with coefficient s≥ 1 and 

degenerated 1+d(x,y) , ‖1 + d(x, y)�‖ ≠ 0�and 𝑥0 ∈ 𝑋. Let 0  ≤ 𝑟 ∈ ℂ and A,B,C and E  are non 

negative reals such that A+√2𝐵 + √2𝐶 + √2𝑠𝐸 < 1. Let S,T:X→X  are mappings satisfying:  

d(Sx,Ty) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐶

𝑑(𝑦,𝑆𝑥)𝑑(𝑥,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐸

𝑑(𝑦,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

1+𝑑(𝑥,𝑦)
���������������������������������������� 

for all x,y∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ . If ‖𝑑(𝑥0, 𝑆𝑥0‖ ≤ (1 − 𝜆)|𝑟| , where λ= max{
𝐴

1−√2𝐵
,

𝐴+√2𝑠𝐸

1−√2𝐵−√2𝑠𝐸
+,                 

then there exist a unique point u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  such that u=Su=Tu. 

Proof: We can prove this result by applying theorem 1.1 by setting D=0. 

Corollary  1.3: Let (X,d) be a complete bi-complex valued b-metric space with coefficient s≥ 1 and 

degenerated 1+d(x,y) , ‖1 + d(x, y)�‖ ≠ 0�and 𝑥0 ∈ 𝑋. Let 0  ≤ 𝑟 ∈ ℂ and A,B,C  be three  non 

negative reals such that  A+√2𝐵 + √2𝐶 < 1. Let S,T:X→X  are mappings satisfying:  

d(Sx,Ty) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐶

𝑑(𝑦,𝑆𝑥)𝑑(𝑥,𝑇𝑦)

1+𝑑(𝑥,𝑦)
��������������������������������������� 

for all x,y∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ . If ‖𝑑(𝑥0, 𝑆𝑥0‖ ≤ (1 − 𝜆)|𝑟| , where λ= 
𝐴

1−√2𝐵
   then there exist a unique point 

u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  such that u=Su=Tu. 

Proof: We can prove this result by applying corollary 1.2 by setting E=0. 

Corollary  1.4: Let (X,d) be a complete bi-complex valued b-metric space with coefficient s≥ 1 and 

degenerated 1+d(x,y) , ‖1 + d(x, y)�‖ ≠ 0�and 𝑥0 ∈ 𝑋. Let 0  ≤ 𝑟 ∈ ℂ and A,B  are non negative 

reals such that  A+√2𝐵 < 1. Let S,T:X→X  are mappings satisfying:  

d(Sx,Ty) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

1+𝑑(𝑥,𝑦)
�����for all x,y∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ . If ‖𝑑(𝑥0, 𝑆𝑥0‖ ≤ (1 − 𝜆)|𝑟| ,  



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where λ= 
𝐴

1−√2𝐵
   then there exist a unique point u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  such that u=Su=Tu. 

Proof: We can prove this result by applying corollary 1.3 by setting C=0.Our result is the extension 

of theorem (3.1) of [6] to the closed ball in complex valued b-metric space. 

Corollary  1.5: Let (X,d) be a complete bicomplex valued metric space with coefficient s≥ 1 and 

𝑥0 ∈ 𝑋. 0≲ 𝑟 ∈ ℂ and A,B,C,D and E are non negative reals such that  

A+√2𝐵 + √2𝐶 + √2𝑠𝐷 + √2𝑠𝐸 < 1. Let T:X→X are mapping satisfying  

d(𝑇𝑛x,�𝑇𝑛y)≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑇𝑛𝑥)𝑑(𝑦,𝑇𝑛𝑦)

1+𝑑(𝑥,𝑦)
+ 𝐶

𝑑(𝑦,𝑇𝑛𝑥)𝑑(𝑥,𝑇𝑛𝑦)

1+𝑑(𝑥,𝑦)
 

������������������������������������������������������������������������������������+𝐷
𝑑(𝑥,𝑇𝑛𝑥)𝑑(𝑥,𝑇𝑛𝑦)

1+𝑑(𝑥,𝑦)
��+ 𝐸

𝑑(𝑦,𝑇𝑛𝑥)𝑑(𝑦,𝑇𝑛𝑦)

1+𝑑(𝑥,𝑦)
                 

for all x,y∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ . If ‖𝑑(𝑥0, 𝑇
𝑛𝑥0‖ ≲𝑖2 (1 − 𝜆)|𝑟| where λ= max{

𝐴+√2𝑠𝐷

1−𝐵−√2𝑠𝐷
,

𝐴+√2𝑠𝐸

1−√2𝐵−√2𝑠𝐸
+,          

Then there exist a unique point u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅  such that u=Tu. 

Proof: For some fixed n, we obtain u∈ B(𝑥0, 𝑟)̅̅ ̅̅ ̅̅ ̅̅ ̅̅ � such that 𝑇𝑛𝑢 = 𝑢.  

The uniqueness follows from 

d(Tu,u)=�𝑑(𝑇𝑇𝑛𝑢, 𝑇𝑛𝑢) ≲𝑖2A�𝑑(𝑇𝑢, 𝑢) + �𝐵
𝑑(𝑇𝑢,𝑇𝑛𝑇𝑢)𝑑(𝑢,𝑇𝑛𝑢)

1+𝑑(𝑇𝑢,𝑢)
+ 𝐶

𝑑(𝑢,𝑇𝑛𝑇𝑢)𝑑(𝑇𝑢,𝑇𝑛𝑢)

1+𝑑(𝑇𝑢,𝑢)
 

                                                                           +𝐷
𝑑(𝑇𝑢,𝑇𝑛𝑇𝑢)𝑑(𝑇𝑢,𝑇𝑛𝑢)

1+𝑑(𝑇𝑢,𝑢)
+ 𝐸

𝑑(𝑢,𝑇𝑛𝑇𝑢)𝑑(𝑢,𝑇𝑛𝑢)

1+𝑑(𝑇𝑢,𝑢)
 

                                               ≲𝑖2A�𝑑(𝑇𝑢, 𝑢) + 𝐶
𝑑(𝑢,𝑇𝑛𝑇𝑢)𝑑(𝑇𝑢,𝑇𝑛𝑢)

1+𝑑(𝑇𝑢,𝑢)
+ 𝐷

𝑑(𝑇𝑢,𝑇𝑛𝑇𝑢)𝑑(𝑇𝑢,𝑇𝑛𝑢)

1+𝑑(𝑇𝑢,𝑢)
 

                                              ≲𝑖2A�𝑑(𝑇𝑢, 𝑢) + 𝐶
𝑑(𝑢,𝑇𝑢)𝑑(𝑇𝑢,𝑢)

1+𝑑(𝑇𝑢,𝑢)
 

Taking norm in above ,we get 

                                ‖𝑑(𝑇𝑢, 𝑢)‖  ≤ A‖𝑑(𝑇𝑢, 𝑢)‖ + 𝐶√2
‖𝑑(𝑢,𝑇𝑢)‖‖𝑑(𝑇𝑢,𝑢)‖

‖1+𝑑(𝑇𝑢,𝑢)‖
 

                     ≤ 𝐴‖𝑑(𝑇𝑢, 𝑢)‖ + 𝐶√2‖𝑑(𝑇𝑢, 𝑢)‖           [as  ‖1 + 𝑑(𝑢, 𝑢)‖ > ‖𝑑(𝑢, 𝑢)‖ ] 

∴  ‖𝑑(𝑇𝑢, 𝑢)‖ ≤ (𝐴 + 𝐶√2)‖𝑑(𝑇𝑢, 𝑢)‖     

This is a contradiction . So u=𝑇𝑛𝑢=Tu . Therefore the fixed point of T  is unique. 

Theorem 2: Let (X,d) be a bi-complex valued complete metric space and T,S:X→X be a self map 

satisfying the following conditions  d(S(x),T(y)) ≲𝑖2 𝛼�max[d(x,y),
𝑑(𝑥,𝑆(𝑥))𝑑(𝑦,𝑇(𝑦))

1+d(Sx,Ty)�
]  ….  (2.1) for all 

x,y∈X, where α is a real with 0 < 𝛼 < 1. Then S and T have a unique common fixed point. 

Proof: Let x∈X be arbitrary. We define a sequence {𝑥𝑛+ in X as follows 𝑥2𝑘+1=S(𝑥2𝑘) and 

𝑥2𝑘+2=T(𝑥2𝑘+1) for k=0,1,2…. 

Then d(x2k+1, x2k+2) = d(S(𝑥2𝑘), T(𝑥2𝑘+1)) 



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                                          ≲𝑖2 𝛼max�,𝑑(𝑥2𝑘, x2k+1),
𝑑(𝑥2𝑘,S(𝑥2𝑘))𝑑(x2k+1,T(𝑥2𝑘+1))

1+𝑑(𝑆𝑥2𝑘,Tx2k+1)
 

∴ d(x2k+1, x2k+2) ≲𝑖2 𝛼d(x2k, x2k+1)                                                                                        (2.2) 

Similarly,  d(x2k+2, x2k+3) = d( T(𝑥2𝑘+1), S(𝑥2𝑘+2))    

                                                  =d(S(𝑥2𝑘+2), T(𝑥2𝑘+1)) 

                                                  ≲𝑖2 �𝛼max�,d(𝑥2𝑘+2, x2k+1),
𝑑(𝑥2𝑘+2,S(𝑥2𝑘+2))𝑑(x2k+1,T(𝑥2𝑘+1))

1+𝑑(𝑆𝑥2𝑘+2,𝑇𝑥2𝑘+1)
] 

                                                  ≲𝑖2 �𝛼d(x2k+2, x2k+1)= 𝛼d(x2k+1, x2k+2)                              (2.3) 

Then from (2.2) and (2.3), we get, 

d(𝑥𝑛+1, 𝑥𝑛+2) ≲𝑖2 𝛼 d(𝑥𝑛, 𝑥𝑛+1) ≲𝑖2 𝛼
2𝑑(𝑥𝑛−1, 𝑥𝑛)……………≲𝑖2 𝛼

𝑛+1d(𝑥0, 𝑥1)     ∀𝑛 ∈ ℕ. 

Now for all m,n∈ℕ we have, 

d(𝑥𝑛, 𝑥𝑚+𝑛) ≲𝑖2 d(𝑥𝑛, 𝑥𝑛+1) + d(𝑥𝑛+1, 𝑥𝑛+2) + ⋯+ d(𝑥𝑚+𝑛−1, 𝑥𝑚+𝑛) 

                       ≲𝑖2 𝛼
𝑛d(𝑥0, 𝑥1) + 𝛼𝑛+1�d(𝑥0, 𝑥1) + ⋯+ 𝛼𝑚+𝑛−1d(𝑥0, 𝑥1) 

∴ d(𝑥𝑛, 𝑥𝑚+𝑛) ≲𝑖2 𝛼
𝑛(1 + 𝛼 + 𝛼2 +⋯+ 𝛼𝑚−1)�d(𝑥0, 𝑥1) 

∴ ‖d(𝑥𝑛, 𝑥𝑚+𝑛)‖ ≤ 𝛼𝑛
1−𝛼𝑚

1−𝛼
�d(𝑥0, 𝑥1)→0 as m,n→∞ 

∴ *𝑥𝑛+�is�a�Cauchy�sequence�in�X. Since�X�is�complete�there�exist�x ∈ X�such�that�𝑥𝑛 → x�as�n

→ ∞. 

Thus lim𝑛→∞ 𝑆(𝑥2𝑛) = lim𝑛→∞ 𝑇(𝑥2𝑛+1) = 𝑥. 

Thus from (2.1), we have d(Sx,x) ≲𝑖2d(Sx,T𝑥2𝑘+1)+d(T𝑥2𝑘+1,x) 

                                                               ≲𝑖2 �𝛼max[d(x, 𝑥2𝑘+1),
𝑑(𝑥,𝑆(𝑥))𝑑(𝑥2𝑘+1,𝑇(𝑥2𝑘+1))

1+d(Sx,𝑇𝑥2𝑘+1)�
] )+d(𝑥2𝑘+2,x)   

                                                               ≲𝑖2 �𝛼d(x, 𝑥2𝑘+1) +d(𝑥2𝑘+2,x)   

∴ ‖d(Sx, x)�‖ ≤ �𝛼‖d(x, 𝑥2𝑘+1)‖ + ‖d(𝑥2𝑘+2, x)��‖    →0 as n→∞. 

    Thus ‖d(Sx, x)‖ = 0. So�𝑆(𝑥) = 𝑥. 

similarly ,  We can prove T(x)=x. 

Thus x is a common fixed point of S and T. 

Now for uniqueness let us assume that 𝑥∗ ∈ 𝑋 is another fixed point of S and T. 

Then d(x, 𝑥∗) = 𝑑(𝑆𝑥, 𝑇𝑥) �≲𝑖2 𝛼max[d(x, 𝑥∗),
𝑑(𝑥,𝑆(𝑥))𝑑(𝑥∗,𝑇(𝑥∗))

1+d(Sx,𝑇𝑥∗)�
] 

                                                   ≲𝑖2 𝛼d(x, 𝑥∗) 

   ⇒(1-𝛼) d(x, 𝑥∗) �≲𝑖2 0 

⇒(1-𝛼)‖d(x, 𝑥∗)�‖ ≤ 0 ⇒ d(x, 𝑥∗) = 0 ⇒ 𝑥 = 𝑥∗. 



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This completes the proof of the theorem. 

Theorem 3: Let (𝑋�, 𝑑�) be a complete bicomplex valued 𝑏�-metric space with the coefficient 𝑠�≥1 

and let 𝑆�,  T: 𝑋� → 𝑋� be mappings satisfying  

d(Sx,Ty) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑆𝑥)𝑑(𝑦,𝑇𝑦)

d�(x,Ty)�+�d�(y,Sx)�+�d�(x,y)
       ………(3.1) 

for all 𝑥�, 𝑦� ∈ 𝑋�, such that x≠ 𝑦 , 𝑑�(𝑥�, 𝑇�𝑦�)+𝑑�(𝑦�, 𝑆�𝑥�)+𝑑�(𝑥�, 𝑦�) ≠ 0, where A,B are nonnegative 

reals with A+ √2𝑠�B < 1 or 𝑑�(𝑆�𝑥�, 𝑇�𝑦�) = 0 if 𝑑�(𝑥�, 𝑇�𝑦�) + 𝑑�(𝑦�, 𝑆�𝑥�) + 𝑑�(𝑥�, 𝑦�) = 0. Then 𝑆� and 

𝑇� have a unique common fixed point. 

Proof: Let 𝑥0�be an arbitrary point in X and define a sequence {𝑥𝑛} in X such that 𝑥2𝑛+1 = 𝑆𝑥2𝑛 and 

𝑥2𝑛+2 = 𝑇𝑥2𝑛+1 where n=0,1,2….  ... … (3.2). Now, we show that the sequence  {𝑥𝑛+ is Cauchy.  

Let x=�x2n and y=x2n+1 in (3.2); we have 

d(x2n+1, x2n+2) = d(Sx2n, Tx2n+1) 

                               ≲𝑖2 𝐴�d(x2n, x2n+1) + 𝐵
d(x2n+1,Tx2n+1)d(x2n,Sx2n)

d(x2n,Tx2n+1)+𝑑(x2n+1,Sx2n)+d(x2n,x2n+1)
             (3.3) 

                                        

                              ≲𝑖2 𝐴�d(x2n, x2n+1) + 𝐵
d(x2n+1,x2n+2)d(x2n,x2n+1)

d(x2n,x2n+2)+𝑑(x2n+1,x2n+1)+d(x2n,x2n+1)
 

This implies that 

‖d(x2n+1, x2n+2)‖ ≤ 𝐴‖d(x2n, x2n+1)‖ 

                                                         +√2𝐵
‖d(x2n+1,x2n+2)‖‖d(x2n,x2n+1)‖

‖d(x2n,x2n+2)‖+‖d(x2n,x2n+1)‖
                                   (3.4) 

As ‖d(x2n+1, x2n+2)‖ ≤ 𝑠(‖d(x2n+1, x2n)‖ + ‖d(x2n, x2n+2)‖)                                      (3.5) 

Therefore ‖d(x2n+1, x2n+2)‖ ≤ 𝐴‖d(x2n, x2n+1)‖ + √2𝑠𝐵‖d(x2n, x2n+1)‖ 

                                                ≤ (𝐴 + √2𝑠𝐵)‖d(x2n, x2n+1)‖                                                 (3.6)                                                                                     

Similarly we get, 

‖d(x2n+2, x2n+3)‖ ≤ (𝐴 + √2𝑠𝐵)�‖d(x2n+1, x2n+2)‖                                                          (3.7) 

Since (𝐴 + √2𝑠𝐵) < 1.� Therefore with  (𝐴 + √2𝑠𝐵) = 𝜆 <1, and for all n≥ 0 ,and consequently,we 

have 

‖d(x2n+1, x2n+2)‖ ≤ �𝜆�‖d(x2n, x2n+1)‖ ≤ � 𝜆�2‖d(x2n−1, x2n)‖ ≤ ⋯ ≤�𝜆�2𝑛+1‖d(x0, x1)‖    (3.8) 

That is,�‖d(xn+1, xn+2)‖ ≤ �𝜆�‖d(xn, xn+1)‖ ≤ � 𝜆�2‖d(xn−1, xn)‖ ≤ ⋯ ≤�𝜆�𝑛+1‖d(x0, x1)‖   (3.9) 

Thus,  for any m> 𝑛 , m, n ∈ ℕ ,we have 

   �‖d(xn, xm)‖ ≤ 𝑠‖d(xn, xn+1)‖ + 𝑠‖d(xn+1, xm)‖         

                           ≤ 𝑠‖d(xn, xn+1)‖ + 𝑠2‖𝑑(𝑥𝑛+1, 𝑥𝑛+2‖ + 𝑠2‖𝑑(𝑥𝑛+2, 𝑥𝑚‖  

                          ≤ 𝑠‖d(xn, xn+1)‖ + 𝑠2‖𝑑(𝑥𝑛+1, 𝑥𝑛+2‖ +⋯+ 𝑠𝑚−𝑛−1‖𝑑(𝑥𝑚−2, 𝑥𝑚−1‖ 



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���������������������������������������������������������������������������������������������������������������������������+𝑠𝑚−𝑛‖𝑑(𝑥𝑚−1, 𝑥𝑚‖     (3.10)      

By using (3.9) we get, 

     �‖d(xn, xm)‖ ≤ 𝑠𝜆𝑛‖𝑑(𝑥0, 𝑥1‖ + 𝑠2𝜆𝑛+1‖𝑑(𝑥0, 𝑥1‖ 

������������������������������������+𝑠3𝜆𝑛+2‖𝑑(𝑥0, 𝑥1‖ +⋯+�𝑠𝑚−𝑛−1𝜆𝑚−2‖𝑑(𝑥0, 𝑥1‖ + 𝑠𝑚−𝑛𝜆𝑚−1‖𝑑(𝑥0, 𝑥1‖             

                             =∑ 𝑠𝑖𝜆𝑛+𝑖−1𝑚−𝑛
𝑖=1 ‖𝑑(𝑥0, 𝑥1‖              

                              ≤
𝑠𝜆𝑛

1−𝜆𝑠
 ‖𝑑(𝑥0, 𝑥1‖ →0 as m,n→∞                                                                     

(3.11) 

This implies that the sequence {𝑥𝑛+�𝑎𝑠�a cauchy sequence in 𝑋.Since X is complete , there exists a 

point u∈ 𝑋 with lim𝑛→∞ 𝑥𝑛 = 𝑢.  

Assume not, then there exists v∈ 𝑋 such that ‖𝑑(𝑢, 𝑆𝑢)‖  =||v||>0                                        (3.12) 

So by using the triangular inequality and (3.1), we get 

v=d(u, Su)�≲𝑖2s�𝑑(𝑢, 𝑥2𝑛+2)+s�𝑑(𝑥2𝑛+2, 𝑆𝑢) 

                      ≲𝑖2 s�𝑑(𝑢, 𝑥2𝑛+2)+s�𝑑(𝑇𝑥2𝑛+1, 𝑆𝑢) 

       ≲𝑖2 s�𝑑(𝑢, 𝑥2𝑛+2)+s�𝐴𝑑(𝑢, 𝑥2𝑛+1)+�𝑠𝐵
d(u,Su)d(x2n+1,Tx2n+1)

d(u,Tx2n+1)+𝑑(x2n+1,Su)+d(u,x2n+1)
           (3.13) 

which implies that 

‖𝑣‖ = �‖𝑑(𝑢, 𝑆𝑢)‖  

≤ 𝑠‖𝑑(𝑢, 𝑥2𝑛+2)‖ + 𝐴𝑠‖𝑑(𝑢, 𝑥2𝑛+1)‖ + 𝑠𝐵√2
‖𝑑(𝑥2𝑛+1,𝑥2𝑛+2)‖‖𝑑(𝑢,𝑆𝑢)‖

‖d(u,Tx2n+1)‖+‖𝑑(𝑥2𝑛+1,𝑆𝑢)‖+‖𝑑(𝑢,𝑥2𝑛+1)‖
      

Taking limit as n→ ∞ ,we get ‖𝑣‖ = �‖𝑑(𝑢, 𝑆𝑢)‖ ≤ 0, a contradiction with (3.12). So ‖𝑣‖ = 0 

Hence ‖𝑑(𝑢, 𝑆𝑢)‖  =0 that is u=Su . 

Similarly ,  we obtain u=Tu. 

For uniqueness assume that 𝑢∗ in 𝑋�is another common fixed point of S and T. Then  

𝑑(𝑢, 𝑢∗) = 𝑑(𝑆𝑢, 𝑇𝑢∗) ≲𝑖2 𝐴�𝑑(𝑢, 𝑢
∗) +

d(u, Su)d(𝑢∗, 𝑇𝑢∗)

d(𝑢, 𝑇𝑢∗) + 𝑑(𝑢∗, 𝑆𝑢) + d(𝑢, 𝑢∗)
 

So that ‖𝑑(𝑢, 𝑢∗)‖ = ‖𝑑(𝑆𝑢, 𝑇𝑢∗)‖ ≤ A‖𝑑(𝑢, 𝑢∗)‖ + 𝐵√2
‖𝑑(𝑢,𝑆𝑢)‖‖𝑑(𝑢∗,𝑇𝑢∗)‖

‖𝑑(𝑢,𝑇𝑢∗)‖+‖𝑑(𝑢∗,𝑆𝑢)‖+‖𝑑(𝑢,𝑢∗)‖
 

                     ≤ 𝐴‖𝑑(𝑢, 𝑢∗)‖  

Hence 𝑢 = 𝑢∗. Therefore u is a unique common fixed point of T and S.  

Now , we consider the second case: 𝑑�(𝑥�, 𝑇�𝑦�) + 𝑑�(𝑦�, 𝑆�𝑥�) + 𝑑�(𝑥�, 𝑦�) = 0. Put x=�x2n and y=x2n+1 

in this expression we get (x2n, 𝑇�x2n+1) + (x2n+1, 𝑆�x2n) + (x2n,�x2n+1) = 0 (for any n) which implies  

(Sx2n, 𝑇�x2n+1)=0 so that x2n = Sx2n= x2n+1 = Tx2n+1= x2n+2. Thus we have x2n = Sx2n= x2n+1, 



Communications on Applied Nonlinear Analysis 

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So there exist K1� and l1 such that K1� =Sl1 = l1 where K1 = x2n+1 and l1 = x2n. Using foregoing 

arguments , one can also show that there exists K2� and l2 such that K2� =Sl2 = l2 where K2 = x2n+2 

and l2 = x2n+1. As (l1, 𝑇�l2) + (l2, 𝑆�l1) + (l1, l2) = 0 (from definition) implies  

𝑑�(Sl1, 𝑇�l2)=0,therefore K1� =Sl1=Tl2 = K2. Thus we obtain that K1� =Sl1=SK1� . similarly ,one can 

also have  TK2 = K2. As K1 = K2 implies SK1 = TK1 = K1, therefore  K1 = K2 is common fixed 

point of S and T. For uniqueness assume that  K1
∗ in X is another common fixed point of S and T. 

Then we have SK1
∗=TK1

∗ =K1
∗  

As  𝑑�(K1, , 𝑇�K1
∗) + 𝑑�(K1

∗, 𝑆�K1, ) + 𝑑�(K1, , K1
∗)= 0, therefore  𝑑�(SK1, TK1

∗
 ) = 𝑑�(K1, , K1

∗)=0. 

This implies that  K1, =K1
∗. 

This completes the proof of the theorem . 

Corollary  3.1: Let (𝑋�, 𝑑�) be a complete bicomplex valued 𝑏�-metric space with the coefficient 𝑠�≥1 

and let   T: 𝑋� → 𝑋� be mappings satisfying  

d(Tx,Ty) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑇𝑥)𝑑(𝑦,𝑇𝑦)

d�(x,Ty)�+�d�(y,Tx)�+�d�(x,y)
      for all 𝑥�, 𝑦� ∈ 𝑋�, such that x≠ 𝑦 ,  

𝑑�(𝑥�, 𝑇�𝑦�)+𝑑�(𝑦�, T𝑥�)+𝑑�(𝑥�, 𝑦�) ≠ 0, where A,B are nonnegative reals with A+ √2𝑠�B < 1 or 𝑑�(T𝑥�, 

𝑇�𝑦�) = 0 if 𝑑�(𝑥�, 𝑇�𝑦�) + 𝑑�(𝑦�, T𝑥�) + 𝑑�(𝑥�, 𝑦�) = 0. Then   has a unique common fixed point. 

Proof: We can prove this result by applying theorem 3 by setting S=T. 

 

Corollary  3.2: Let (𝑋�, 𝑑�) be a complete bicomplex valued 𝑏�-metric space with the coefficient 𝑠�≥1 

and let   T: 𝑋� → 𝑋� be mappings satisfying (for some fixed n) 

d(𝑇𝑛x,�𝑇𝑛y) ≲𝑖2Ad(x,y)+B
𝑑(𝑥,𝑇𝑛𝑥)𝑑(𝑦,𝑇𝑛𝑦)

d�(x,𝑇𝑛y)�+�d�(y,𝑇𝑛x)�+�d�(x,y)
      for all 𝑥�, 𝑦� ∈ 𝑋�, such that x≠ 𝑦 ,  

𝑑�(𝑥�, 𝑇𝑛𝑦�)+𝑑�(𝑦�, 𝑇𝑛𝑥�)+𝑑�(𝑥�, 𝑦�) ≠ 0, where A,B are nonnegative reals with A+ √2𝑠�B < 1 or 

𝑑�(𝑇𝑛𝑥�, 𝑇𝑛𝑦�) = 0 if 𝑑�(𝑥�, 𝑇𝑛𝑦�) + 𝑑�(𝑦�, 𝑇𝑛𝑥�) + 𝑑�(𝑥�, 𝑦�) = 0. Then   has a unique common fixed 

point. 

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