Question

show that f(z)=1/z is analytic or not

Answer 1

Answer:

Whenever f ′ (z0) exists, f is said to be analytic (or regular, or holomorphic) at

the point z0. The function is analytic throughout a region in the complex plane

if f ′  exists for every point in that region. Any point at which f ′ does not exist

is called a singularity or singular point of the function f.

If f(z) is analytic everywhere in the complex plane, it is called entire.

1/z is analytic except at z = 0, so the function is singular at that point.

Step-by-step explanation:

Answer 2

Ques. Check whether the function f (z)=1/z is analytic or not using the concept of complex analysis

The function f(z) = 1/z is not analytic at z = 0.

To see this, we can use the definition of analyticity, which states that a function is analytic at a point if it has a derivative at that point, and if this derivative is continuous in a neighborhood of the point.

Let's compute the derivative of f(z) = 1/z at z = 0:

f'(z) = -1/z^2

The derivative of f(z) at z=0 is not defined, because 1/0 is undefined. Therefore, f(z) is not analytic at z=0.

Alternatively, we can show that f(z) does not satisfy the Cauchy-Riemann equations, which are necessary conditions for a function to be analytic. The Cauchy-Riemann equations state that if a function f(z) = u(x,y) + i*v(x,y) is analytic, then its partial derivatives satisfy:

∂u/∂x = ∂v/∂y and ∂u/∂y = -∂v/∂x

Let's apply these equations to f(z) = 1/z:

f(z) = 1/z = x/(x^2+y^2) - i*y/(x^2+y^2)

∂u/∂x = 1/(x^2+y^2) - (2x^2)/(x^2+y^2)^2

∂u/∂y = (-2xy)/(x^2+y^2)^2

∂v/∂x = (2xy)/(x^2+y^2)^2

∂v/∂y = -1/(x^2+y^2) + (2y^2)/(x^2+y^2)^2

We can see that these partial derivatives do not satisfy the Cauchy-Riemann equations, except at points where x=0 or y=0, which corresponds to the real and imaginary axis respectively. At z=0, both the partial derivatives are undefined, so the Cauchy-Riemann equations do not apply. Therefore, f(z) is not analytic at z=0.

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