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The aim of this paper is to prove common fixed point theorems for variants of weak compatible maps in a complex valued-metric space. In this paper, we generalize various known results in the literature using (CLRg) property. The concept of (CLRg) does not require a more natural condition of closeness of range.

Recently, Azam et al. [

Let ℂ be the set of complex numbers and z_{1}, z_{2} ∊ ℂ, recall a natural partial order relation ≾ on ℂ as follows:

Definition 2.1. [

(C1)

(C2)

(C3)

Then d is called a complex-valued metric on X, and (X, d) is called a complex-valued metric space.

Example 2.1. [

Definition 2.2. [_{n}} sequence is i) convergent if for every

ii) a Cauchy sequence, if for every

Definition 2.3. [

Example 2.2. [

where a is any real constant. Then (X, d) is a complex-valued metric space. Suppose self maps f and g be defined as:

and

Clearly, f and g are weakly compatible self maps.

In 2011, Sintunavarat and Kumam [_{g}) property, defined as:

Definition 2.4. A pair (f, g) of self-mappings is said to be satisfy the common limit in the range of g property if there exists a sequence

Definition 3.1. Let (X, d) be a complex valued metric space and (f, g) be a pair of self mappings on X and

and define the following conditions:

A) For arbitrary

such that

B) For arbitrary

such that

C) For arbitrary

such that

Conditions A), B) and C) are called strict contractive conditions.

Theorem 3.1. Let f and g be two weakly compatible self mappings of a complex valued metric space (X, d) such that

(3.1) f, g satisfy (CLRg) property;

(3.2) for all

such that

Then f and g have a unique common fixed point in X.

Proof. Since f and g satisfy the (CLRg) property, there exists a sequence {x_{n}} in X such that

We first show that fx = gx. Suppose not, i.e., fx ≠ gx.

From (3.2),

where

Three cases arises:

i) If

then (3.3) implies

Taking limit as n→∞,

which gives,

ii) If

then (3.3) implies,

Taking limit as n→∞,

i.e.,

iii) If

Making limit as n→∞,

i.e.,

Hence, from all three cases, gx = fx.

Now let z = fx = gx. Since f and g are weakly compatible mappings fgx = gfx which implies that fz = fgx = gfx = gz.

We claim that fz = z. Let, if possible, fz ≠ z.

Now

where

Two cases arises:

i) If

which gives,

ii) If

which gives,

Hence, from two cases, it is clear that fz = z = gz.

Hence z is a common fixed point of f and g.

For uniqueness, suppose that w is another common fixed point of f and g.

We shall prove that z = w. Let, if possible, z ≠ w.

Then

where

Again, two possible cases i) If

which gives,

ii) If

which gives,

Hence, z = w.

So, we can say that f and g have a unique common fixed point.

Remark 3.1. Theorem 3.1 also holds true if

Definition 3.2. Let (X, d) be a complex valued metric space, and let f, g: X→X. Then f is called a g-quasicontraction, if for some constant

such that

Theorem 3.2. Let f and g be two weakly compatible self mappings of a complex valued metric space (X, d) such that

(3.6) f is a g-quasi-contraction;

(3.7) f and g satisfy (CLRg) property.

Then f and g have a unique common fixed point.

Proof. Since f and g satisfy the (CLRg) property, there exists a sequence {x_{n}} in X such that

We first claim that fx = gx. Suppose not. Since, f is a g-quasi-contraction, therefore

for some

Following five cases arises:

i) If

taking limit as n→∞,we have

which gives,

ii) If

taking limit as n→∞,we have

which gives,

iii) If

taking limit as n→∞, we have

which gives,

iv) If

taking limit as n→∞,we have

which gives,

v) If

taking limit as n→∞,we have

which gives,

Thus from all fives possible cases, gx = fx.

Now, let z = fx = gx. Since f and g are weakly compatible mappings fgx = gfx which implies that fz = fgx = gfx = gz.

We claim that fz = z. Suppose not, then by (3.6), we have

where

Two cases arises:

i) If

which gives,

ii) If

which gives,

Thus, fz = z = gz.

Hence, z is a common fixed point of f and g.

For uniqueness, suppose that w is another common fixed point of f and g in X.

By (3.6), we have

where

Two possible cases arises:

i) If

which gives

ii) If

which gives

Hence, z = w i.e., f and g have a unique common fixed point.