Removable Singularity is a term which is related with the field of complex analysis and is linked with all types of holomorphic function. In simple words, removable singularities are those points on the graph of the function at which the holomorphic function always comes out to be undefined, and therefore, we can always redefine the function in those ways by which the function can become regular around a specific neighborhood of the point which makes the function undefined.

In order to find the removable singularities of any given function, the first thing which we need to know is about the definition of singularity and what are its specific types in the field of complex analysis. So, any singularity, which is also know as isolated or a point of a function is defined as follows:

If f be any function which is defined on a set that is open, then if the function f is not defined at a point z_{0}, and is then redefined on a deleted neighborhood (which means is defined in some specific neighborhood of that point, deleting that point), such that 0 < $\left | z-z_{0} \right |$ < r, this means that the point z_{0} is the point of singularity of the function f.

Lets consider a function f (z) = $\frac{1}{z}$, which is defined on C - 0. Then the point z = 0 will be considered as the

point of singularity of the function f, because of the fact that the function f becomes undefined at the point 0, and can be defined at any other point except 0. Also, we will find that $\left | f (z) \rightarrow infinity \right |$ as the point z $\rightarrow $ 0. We can always generalize this result to get that, the function f (z) = 1 / z^{n}, where n is always greater than equal to one, that is a positive integer, will have z = 0 as a point of singularity and $\left | f (z) \rightarrow infinity \right |$ as z $\rightarrow $ 0. Lets consider another function as f (z) = e^{1/z}. Here, it is literally impossible to define f (0) so as to make the function f as

continuous at the point 0, because of the fact that e^$\frac{1}{z}$ will take each non zero number to infinity near the

neighbourhood of 0. Moreover, we can also find out that |f (z)| will not tend to infinity as z approaches near zero. Hence, this function helps in understanding how to find out the removable singularities of a function, by first understanding about the zeroes of the holomorphic functions, because all the singularities of 1/z^{n} comes out on those points in which the denominator z^{n} has zeroes in it.

The next important thing to find out the removable singularities is by finding out the poles of the holomorphic functions, which are defined as follows:

Let f be any holomorphic function with the point of singularity at z_{0}, and if f is a non zero function defined in some disc which is centered at z_{0}, then the function $\frac{1}{f}$ is holomorphic in that disc. Now lets define, 1 / f at the point z_{0}, as equal to

zero, then the function f has a pole of order n if $\frac{1}{f}$ is holomorphic at the point z_{0}, which the zero of order n at the point z_{0}.

Also, if the point z_{0} is a pole of order 1, for the function f, then z_{0} is often referred to as a simple pole of the function f.

Let f (z) be any function at the point z and consider z_{0} as a singularity of this function, and consider it redefining the complete function as f (z_{0}) in such a manner, so that the function f is holomorphic at the point of singularity z_{0}, then we define this point, z_{0} as a point of removable singularity of the function f. Thus, in simple words, we can always say that removable singularities are those points of the holomorphic functions which are made by the points in the domain of the function which has to be deleted, because at those points the function itself become undefined.

**Some of the examples of the removable singularities are explained as below:**

**Example 1:** Lets consider the very basic example of a function f (z) = z / z, which is defined on C – 0, then this function will just represent a constant function f (z) = 1, which is defined on C – 0. Again, then the point z = 0 will be considered as the point of singularity of the function f, because of the fact that the function f becomes undefined at the point 0, and can be defined at any other point except 0. We can also redefine f (0) = 1 to make the function f (z) both continuous and holomorphic too at the point z = 0.

**Example 2:** Find the removable singularity of $f (z)$ = $\frac{sin^2 x}{ x}

.

**Solution:** By applying the definition of limits, we can expand the function sin^2 x / x as:

$\frac{sin^2 x}{ x}$ = $\frac{sin x }{ x }$ $\times sin x$. Now by applying limits as x tends to zero, we get:

$\frac{sin^2 x}{ x}$ = $Lim_{x \rightarrow 0}$$\frac{sin x }{ x }$ $\times sin x$ = 1 $\times$ 0 = 0, which means

that this function has a removable singularity with the pole at x = 0.

Example 3: How would we remove the singularity of the function $f (z)$ = $\frac{sin z }{ z}$. Give explanation.

**Solution: **We know that the function $f (z)$ = $\frac{sin z }{ z}$ is analytic in the entire punctured disc of z such that 0 < |z| < R. But, the quotient of this function is not defined only at one single point of z = 0, therefore, we can write the function sin z = z + z^{3} g(z), where g is an entire function, because of the fact that the function sin z in an entire function too.

This means that for z not equal to zero, we can redefine the function as:

f (z) = $\frac{sin z }{ z}$ = 1 + z^{2} g (z).

Also note that the function 1 + z^{2} g (z) is also an entire function, therefore, we have defined f (0) = 1, which shows f is now analytic at z = 0 also, and thus, in this way, the singularity has been removed now.

**Example 4**: Let z_{0} is a singularity of the function f, and limit as z approaches towards z_{0} of [(z- z_{0}) * f (z)] = 0, then show that z_{0} will a removable singularity of the function f.

**Solution:** Lets define a function g (z) = [(z- z_{0}) * f (z)] where it is assumed that z is not equal to z_{0}, and g (z_{0}) = 0. Then, g (z) would be continuous at z_{0} and also holomorphic at z_{0}. Therefore, let’s define h (z) = [g (z) – g (z_{0})] / [z – z_{0}], where z is not equal to z_{0} and the first derivative of g (z) otherwise. Then, we can say, h is holomorphic at the point z_{0} and also at z not equal to z_{0}, hence,

h (z) = (z – z_{0}) f (z) / (z – z_{0}) = f(z). Therefore, we can now say that h (z) = f (z) with h ( z_{0}) defined in such a manner as to mold the function f (z) holomorphic at the point z_{0}, and hence, z_{0} is a removable singularity of the function f.

If f be any function which is defined on a set that is open, then if the function f is not defined at a point z

Lets consider a function f (z) = $\frac{1}{z}$, which is defined on C - 0. Then the point z = 0 will be considered as the

point of singularity of the function f, because of the fact that the function f becomes undefined at the point 0, and can be defined at any other point except 0. Also, we will find that $\left | f (z) \rightarrow infinity \right |$ as the point z $\rightarrow $ 0. We can always generalize this result to get that, the function f (z) = 1 / z

continuous at the point 0, because of the fact that e^$\frac{1}{z}$ will take each non zero number to infinity near the

neighbourhood of 0. Moreover, we can also find out that |f (z)| will not tend to infinity as z approaches near zero. Hence, this function helps in understanding how to find out the removable singularities of a function, by first understanding about the zeroes of the holomorphic functions, because all the singularities of 1/z

The next important thing to find out the removable singularities is by finding out the poles of the holomorphic functions, which are defined as follows:

Let f be any holomorphic function with the point of singularity at z

zero, then the function f has a pole of order n if $\frac{1}{f}$ is holomorphic at the point z

Also, if the point z

Let f (z) be any function at the point z and consider z

.

$\frac{sin^2 x}{ x}$ = $\frac{sin x }{ x }$ $\times sin x$. Now by applying limits as x tends to zero, we get:

$\frac{sin^2 x}{ x}$ = $Lim_{x \rightarrow 0}$$\frac{sin x }{ x }$ $\times sin x$ = 1 $\times$ 0 = 0, which means

that this function has a removable singularity with the pole at x = 0.

Example 3:

This means that for z not equal to zero, we can redefine the function as:

f (z) = $\frac{sin z }{ z}$ = 1 + z

Also note that the function 1 + z

h (z) = (z – z

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