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Where is the function f defined by f(z)=3x2y-y3+x2-3x+i(-x3+3xy2+2y2) differentiable? Where is it analytic?

Last Updated : 30 May, 2023
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The number of the form x+ iy where x and y are real numbers and ‘i’ is iota  (or √-1), ‘i’ or iota is the solution for the equation a2+1=0 is called as complex numbers. A complex number consists of two parts, the real part, and an imaginary part. In x+iy, x is called the real part and iy is the imaginary part.

Example of Complex numbers: 3+i4, -2i,etc. 

Complex Functions

The function of a complex variable is a relation that maps a complex number say ‘z’ to another complex number ‘w’.And it is denoted as,

w=F(z)

If z=x+iy and w=U+ iV, generally U and V are implicit functions of x and y. And in general ‘w’ can be denoted as,

w=U(x,y)+ iV(x,y)

Examples of complex functions,

  • W = z3
  • W = x2-y2+5xy+ i(x+y)

We can see in the second example that the real part and imaginary part of ‘W’ can be the implicit function of x and y where z=x+iy.

Note: Implicit function is defined as a function with possible multiple variables.

Example: 

  • G(x,y)=x2+y2
  •  G(w,t)=5wt

Differentiable complex Function

A complex function F[z] is said to be differentiable at z=z0 if the following limit exists at z=z0.

lim δz⇢0 (F(z+δz)-F(z))/δz  

exists at z=z0

In the case of real functions {say F(x)}, we used to approach the limit from two directions only i.e., Left-hand limit(L.H.L) and Right-Hand Limit (R.H.L)and we used to say if L.H.L=R.H.L = F(x) then the function is differentiable But in case of complex functions there are an infinite number of directions from which the limit can be approached.

For a better understanding of solving the above limit let’s consider the following example:

Example: Check Differentiability of F(z)=z2.

Solution:

We know that for a function to be differentiable at any z0 the following limit should exist at z=z

lim δz⇢0 (F(z+δz)-F(z))/δz  exists at z=z0

Now for the function to be differentiable the limit value along any two directions(path) must be equal.

F(z+δz)-F(z)=(z+δz)2-z2 =z2 +2zδz+δz2 -z2= 2zδz+ δz2

lim δz⇢0 (F(z+δz)-F(z))/δz 

⇒lim δz⇢0 (2zδz+ δz2)/δz

We have z=x+ iy and δ z=δ x + iδ y.
Consider first path along horizontal line(parallel to x-axis) i.e, δy=0

lim δx⇢0 (δx +(2x+2iy) = 2x+2iy =2z

Consider second path along vertical line(parallel to y axis) i.e., δ x=0.

lim δy⇢0 (δx +(2x+2iy) = 2x+2iy =2z

So the limit values along horizontal and vertical direction are both equal to 2(x+iy) or 2z.
And as the function consists of polynomial parameters it is continuous also, Hence the function F(z)=z2 is differentiable at all z ∈ complex number.

Note: If F(z)=zn  then F'(z)=nz{n-1}.

Some basic Rules in Differentiation of complex functions:

  • dc/dz = 0  where c is a complex constant
  • d (f ± g) /dz= df/dz ± dg/dz
  • d(cf )/dz = c df/dz  where c is a complex constant
  • d(f.g)/dz = g df/dz + f dg/dz
  • d/dz ( f/g ) = (g df/dz −f dg/dz)/g2

Note: In the above mentioned rules ‘c’ is complex constant ,and ‘g’ and ‘f’ are complex functions.

Sufficient conditions  for a complex Function to be  differentiable 

A function F[z]=U(x,y)+iV(x,y) is said to be differentiable in region R of a complex plane if the following conditions are true for it:

  1. Partial derivatives of U(x,y) and V(x,y) w.r.t x and y must exist and be continuous.
  2. Partial derivatives of U(x,y) w.r.t x =  Partial derivatives of V(x,y) w.r.t y
  3. Partial derivatives of U(x,y) w.r.t y =  (-1 ) × Partial derivatives of V(x,y) w.r.t x

Theorem: A complex function F[z]=U(x,y)+iV(x,y) is said to be differentiable at z0 if the first order derivatives of U and V w.r.t x and y exist and also satisfy the Cauchy-Riemann Equations.

Consider F(z)= U(x,y)+ i V(x,y) where z=x+ iy, then Cauchy Riemann equations are given by:

  1. (δU/δx) = (δV/δy)
  2. (δU/δy) = −(δV/δx)

Working on Cauchy-Riemann Equations

Till now we have seen that for a function F(z) to be differentiable at z0 the given limit should exist

 lim δz⇢0 (F(z+δz)-F(z))/δz  exists at z=z0

Now F(z)=U(x,y) +iV(x,y)

And so F(z+δz)=U(x+δx,y+δy) + V(x+δx,y+δy)

Therefore, 
∴ {F (z + δz) − F (z)}/δz = {U (x + δx, y + δy) + iV (x + δx, y + δy) − U (x, y) − iV (x, y)}/δx + iδy                          ⇢(A)

As (z=x+iy and δz=δx+iδy)

Now as per differentiability as the limit exists, the Limit value along any two directions(path) must be equal.
Consider first path along horizontal line(parallel to x-axis) i.e, δy=0.
So putting δy=0 in equation (A) we get

F ′(z) = { U (x + δx, y) + iV (x + δx, y) − U (x, y) − iV (x, y) }/δx
F ′(z) ={ U (x + δx, y) − U (x, y) }/δx + i(V (x + δx, y) − V (x, y))/δx
F ′(z) = δU/δx + i δV/δx → (B)

Consider second path along vertical line(parallel to y axis) i.e., δx = 0, we get

F ′(z) = { U (x, y + δy) + iV (x, y + δy) − U (x, y) − iV (x, y) }/iδy
F ′(z) =  { U (x, y + δy) − U (x, y) }/iδy + i(V (x, y + δy) − V (x, y))/iδy
F ′(z) = (1/i) δU/δy + δV/δy

Since (1/i = i/(i2) =-i) we have,

F'(z)= -i(δU/δy) +(δV/δy)  -> (C)

As we know differentiation of a complex function from any direction is equal. Therefore, From (B) and (C) we have

F'(z)= -i(δU/δy) +(δV/δy)= (δU/δx) +i(δV/δx)

Equating the real parts we have
(δU/δx) = (δV/δy)
Equating the imaginary part we have
δU/δy = −(δV/δx)

Hence this is how we arrive at Cauchy Riemann Equations.

Analyticity

A complex function F(z) is said to be analytic at z = z0 if it is differentiable at z = z0 and also in some neighborhoods of z0.
Neighborhood: The neighborhood of a point z0 is a set of all points z where |z − z0| < k where k is a positive constant.
If a complex function is analytic in a particular domain that means it is analytic at every single point in that domain. If a complex function is entirely analytic (for all z in the complex plane ) then the function is known as the Entire Function.
Note: Polynomial complex functions of z are entire functions. An example is given below:
Example F (z) = z3 + 3z2 − 6z will be analytic everywhere.
A crucial point to note is that if a function is analytic in a region then it will be definitely differentiable in that same region but vice-versa may not always be true. An analytic function will satisfy the Cauchy Riemann Equations.
Let us see how we check for analyticity with an example:

Example: Discuss the analyticity of  F (z) = ex(cosy + i siny)

Solution:

We have U (x, y) = excosy and V (x, y) = exsiny
δU/δx = excosy
δU/δy = −exsiny
δV/δx = exsiny
δV/δy = excosy
So we can clearly see that cauchy reimann equations are satisfied
δU/δx = δV/δy = excosy
δU/δy = − δV/δx = exsiny

And Cauchy-Reimann equations for the function is true for all values of x and y and so the function is analytic for all z in complex plane.

Some important properties of Analytic Function:

  1. If complex functions F1 and F2 are analytic then F1 ± F2, F1 ∗ F2, and F1/F2 are also analytic whereas for F1/F2 we have F2 is not equal to 0.
  2. If a function is an Entire Function it implies it is analytic, further it implies differentiability which again implies continuity.
    Entire ⇒ Analytic ⇒ Differentiable ⇒ Continuity.

Where is the function f defined by f(z)=3x2y − y3 + x2 − 3x + i(−x3 + 3xy2 + 2y2) differentiable? Where is it analytic?

Answer:

F (z) = U (x, y) + iV (x, y)
F (z) = 3x2y − y3 + x2 − 3x + i(−x3 + 3xy2 + 2y2)
U (x, y) = 3x2y − y3 + x2 − 3x and V (x, y) = −x3 + 3xy2 + 2y2
As both U and V are polynomial functions the partial derivatives of U and V w.r.t x and y will exist and be continuous.
Partial derivatives of U(x,y) w.r.t x= δU/δx = 6xy + 2x − 3
Partial derivatives of V(x,y) w.r.t x= δV/δx = 3y2 − 3x2
Partial derivatives of U(x,y) w.r.t y= δU/δy = 3x2 − 3y2
Partial derivatives of V(x,y) w.r.t y= δV/δy = 6xy + 4y

Partial derivatives of U and V w.r.t x and y are polynomial functions so it exists and are continuous everywhere.
Now according to Cauchy Reimann Equations we have
δU/δx = δV/δy
δU/δy = − δV/δx
Equating δU/δx and δV/δy we get
2x − 3 = 4y
Equating δU/δy and − δV/δx we can clearly see that
δU/δy = − δV/δx = 3x2 − 3y2
Therefore for all z = x + iy having the values of x and y satisfying the linear condition 2x − 3 = 4y the function F (z) will be differentiable.
F (z) is differentiable only along the line 2x − 3 = 4y , not through any region (neighborhood)R. Hence F (z) is nowhere analytic.

Sample Problems

Problem 1: Prove that the greatest integer function defined by f(x) = [x] , 4 < x < 8 is not differentiable at x = 5 and x = 6.

Solution:

As the question given  f(x) = [x] where x is greater than 4 and less than 8. So we have to check the function is differentiable at point x =5 and at x = 6 or not. To check the differentiability of function, as we discussed above in Differentiation that LHD at(x=a) = RHD at (x=a) which means,

Lf’ at (x = a) = Rf’ at (x = a) if they are not equal after solving and putting the value of a in place of x then our function should not differentiable and if they both comes equal then we can say that the function is differentiable at x = a, we have to solve for two points x = 5 and x = 6. Now, let’s solve for x = 5

f(x) = [x]

Put x = 5 + h

Rf’ = limh -> 0 f(5 + h) – f(5)

= limh -> 0 [5 + h] – [5]

Since [h + 5] = 5

= limh -> 0 (5 – 5) / h = 0

Lf'(1) = limh -> 0 [f(5 – h) – f(5)] / -h

= limh -> 0 ( [5 – h] – [5] ) / -h

Since [5 – h] = 4

= limh -> 0 (4 – 5) / -h

= -1 / -0

= ∞

From the above solution it is seen that Rf’ ≠ Lf’, so function f(x) = [x] is not differentiable at x = 5. Now, let’s check for x = 6. As we solved for x = 5 in the same we are going to solve for x = 6. The condition should be the same we have to check that, Lf’ at (x = 6) = Rf’ at (x = 6) or not if they are equal then our function is differentiable at x = 6 and if they are not equal our function is not differentiable at x = 6. So, let’s solve.

f(x) = [x]

Differentiability at x=6

Put x = 6+h

Rf'(6) = limh -> 0 f(6 + h) – f(6)

= limh -> 0 ([6 + h] – [6]) / h

Since, 6 + h = 6

= limh -> 0 (6 – 6) / h

= limh ->0 0 / h

=0

Lf'(6) = limh -> 0 (f(6 – h) – f(6)) / -h

= limh -> 0 ([6 – h] – [6]) / -h

= limh -> 0 (5 – 6) / -h

Since [5 – h] =6

= -1 / -0

= ∞

From the above solution it is seen that Rf'(2) ≠ Lf'(2), so f(x) = [x] is not differentiable at x = [6]

Problem 2: f(x) = \left\{\begin{matrix} x\times \frac{e^{1/x} + 1, x \neq 0}{e^{1/x} - 1, x = 0} \end{matrix}\right. Show that the above function is not derivable at x = 0.

Solution:

As we know to check the differentiability we have to find out Lf’ and Rf’ then after comparing them we get to know whether the function is differentiable at the given point or not. So let’s first find the Rf'(0).

Rf'(0) = limh -> 0 f(0 + h) – f(0)

= limh -> 0 (f(h) – f(0)) / h

= limh -> 0 h . [{(e(1 / h) + 1) / (e(1 / h) – 1) } – 0]/h

= limh -> 0 (e(1 / h) + 1) / (e(1 / h) – 1)

Multiply by e(1 / h)

= limh -> 0 {1 + e(1 / h) / 1 – e(1 / h)}

= (1 + 0) / (1 – 0)

= 1

After solving we had find the value of Rf'(0) is 1. Now after this let’s find out the Lf'(0) and then we will check whether the function is differentiable or not.

Lf'(0) = limh -> 0 { f(0 – h) – f(0) } / -h

= limh -> 0 -h . [{e(-1 / h) + 1 / e(-1 / h) – 1} – 0] / -h

= limh -> 0 { (e(-1 / h) + 1) / (e(-1 / h) – 1) }

= limh -> 0 { (1 + e(-∞)) / (1-e(-∞))}

= (0 + 1) / (0 – 1)

= -1  

As we saw after solving Lf'(0) the value we get -1. Now checking if the function is differentiable or not, Rf'(0) ≠ Lf'(0) (-1≠1). Since Rf'(0) ≠ Lf'(0), so f(x) is not differentiable at x = 0.

Problem 3: A function is f(x) defined by

f(x) = 1 + x of x < 3

f(x) = 5 – x of x ≥ 3
 

Check, If function f(x) differentiable at x = 3?

Solution:

So, for finding Lf'(3) we take the function f(x) = 1 + x, in the same way for finding Rf'(3) we take the function f(x) = (5 – x). Let’s find out Lf'(2) and Rf'(2)

Lf'(3) = limh -> 0 {f(3 – h) – f(3)} / -h

= limh -> 0 [[(2 – h) + 1] – [5 – 2]] / -h

= limh -> 0 (3 – h – 3) / -h

= limh -> 0 -h / h

= -1

Rf'(3) = limh -> 0 {f(3 + h) – f(3)} / h

= limh -> 0 [[5 – (3 + h)] – 2] / h

=limh -> 0 h / h

= 1  

In the first line Lf'(3) after putting in the formula, for f(3) we are putting the second function (5 – x). After solving the Lf'(3) we get the value -1. For calculating Rf'(3) we are using the second function 5-x and putting in the formula of Rf’, on solving the Rf'(3) we get the value 1. Since, Rf'(3) ≠ Lf'(3) so we can say the function f(x) is not differentiable at x = 3.



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