# 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

The number of the form x+ iy where x and y are real numbers and ‘i’ is iota  (or âˆš-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 Î´zâ‡¢0 (F(z+Î´z)-F(z))/Î´z

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 Î´zâ‡¢0 (F(z+Î´z)-F(z))/Î´z  exists at z=z0

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

F(z+Î´z)-F(z)=(z+Î´z)2-z2 =z2 +2zÎ´z+Î´z2 -z2= 2zÎ´z+ Î´z2

lim Î´zâ‡¢0 (F(z+Î´z)-F(z))/Î´z

â‡’lim Î´zâ‡¢0 (2zÎ´z+ Î´z2)/Î´z

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

lim Î´xâ‡¢0 (Î´x +(2x+2iy) = 2x+2iy =2z

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

lim Î´yâ‡¢0 (Î´x +(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 âˆˆ 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 Â± g) /dz= df/dz Â± 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 âˆ’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 ) Ã— 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. (Î´U/Î´x) = (Î´V/Î´y)
2. (Î´U/Î´y) = âˆ’(Î´V/Î´x)

### 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 Î´zâ‡¢0 (F(z+Î´z)-F(z))/Î´z  exists at z=z0

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

And so F(z+Î´z)=U(x+Î´x,y+Î´y) + V(x+Î´x,y+Î´y)

Therefore,
âˆ´ {F (z + Î´z) âˆ’ F (z)}/Î´z = {U (x + Î´x, y + Î´y) + iV (x + Î´x, y + Î´y) âˆ’ U (x, y) âˆ’ iV (x, y)}/Î´x + iÎ´y                          â‡¢(A)

As (z=x+iy and Î´z=Î´x+iÎ´y)

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, Î´y=0.
So putting Î´y=0 in equation (A) we get

F â€²(z) = { U (x + Î´x, y) + iV (x + Î´x, y) âˆ’ U (x, y) âˆ’ iV (x, y) }/Î´x
F â€²(z) ={ U (x + Î´x, y) âˆ’ U (x, y) }/Î´x + i(V (x + Î´x, y) âˆ’ V (x, y))/Î´x
F â€²(z) = Î´U/Î´x + i Î´V/Î´x â†’ (B)

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

F â€²(z) = { U (x, y + Î´y) + iV (x, y + Î´y) âˆ’ U (x, y) âˆ’ iV (x, y) }/iÎ´y
F â€²(z) =  { U (x, y + Î´y) âˆ’ U (x, y) }/iÎ´y + i(V (x, y + Î´y) âˆ’ V (x, y))/iÎ´y
F â€²(z) = (1/i) Î´U/Î´y + Î´V/Î´y

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

F'(z)= -i(Î´U/Î´y) +(Î´V/Î´y)  -> (C)

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

F'(z)= -i(Î´U/Î´y) +(Î´V/Î´y)= (Î´U/Î´x) +i(Î´V/Î´x)

Equating the real parts we have
(Î´U/Î´x) = (Î´V/Î´y)
Equating the imaginary part we have
Î´U/Î´y = âˆ’(Î´V/Î´x)

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 âˆ’ 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 âˆ’ 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
Î´U/Î´x = excosy
Î´U/Î´y = âˆ’exsiny
Î´V/Î´x = exsiny
Î´V/Î´y = excosy
So we can clearly see that cauchy reimann equations are satisfied
Î´U/Î´x = Î´V/Î´y = excosy
Î´U/Î´y = âˆ’ Î´V/Î´x = 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 Â± F2, F1 âˆ— 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 â‡’ Analytic â‡’ Differentiable â‡’ Continuity.

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

F (z) = U (x, y) + iV (x, y)
F (z) = 3x2y âˆ’ y3 + x2 âˆ’ 3x + i(âˆ’x3 + 3xy2 + 2y2)
U (x, y) = 3x2y âˆ’ y3 + x2 âˆ’ 3x and V (x, y) = âˆ’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= Î´U/Î´x = 6xy + 2x âˆ’ 3
Partial derivatives of V(x,y) w.r.t x= Î´V/Î´x = 3y2 âˆ’ 3x2
Partial derivatives of U(x,y) w.r.t y= Î´U/Î´y = 3x2 âˆ’ 3y2
Partial derivatives of V(x,y) w.r.t y= Î´V/Î´y = 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
Î´U/Î´x = Î´V/Î´y
Î´U/Î´y = âˆ’ Î´V/Î´x
Equating Î´U/Î´x and Î´V/Î´y we get
2x âˆ’ 3 = 4y
Equating Î´U/Î´y and âˆ’ Î´V/Î´x we can clearly see that
Î´U/Î´y = âˆ’ Î´V/Î´x = 3x2 âˆ’ 3y2
Therefore for all z = x + iy having the values of x and y satisfying the linear condition 2x âˆ’ 3 = 4y the function F (z) will be differentiable.
F (z) is differentiable only along the line 2x âˆ’ 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,

Lfâ€™ at (x = a) = Rfâ€™ 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, letâ€™s solve for x = 5

f(x) = [x]

Put x = 5 + h

Rfâ€™ = limh -> 0 f(5 + h) â€“ f(5)

= limh -> 0 [5 + h] â€“ [5]

Since [h + 5] = 5

= limh -> 0 (5 â€“ 5) / h = 0

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

= limh -> 0 ( [5 â€“ h] â€“ [5] ) / -h

Since [5 â€“ h] = 4

= limh -> 0 (4 â€“ 5) / -h

= -1 / -0

= âˆž

From the above solution it is seen that Rfâ€™ â‰  Lfâ€™, so function f(x) = [x] is not differentiable at x = 5. Now, letâ€™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, Lfâ€™ at (x = 6) = Rfâ€™ 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, letâ€™s solve.

f(x) = [x]

Differentiability at x=6

Put x = 6+h

Rf'(6) = limh -> 0 f(6 + h) â€“ f(6)

= limh -> 0 ([6 + h] â€“ [6]) / h

Since, 6 + h = 6

= limh -> 0 (6 – 6) / h

= limh ->0 0 / h

=0

Lf'(6) = limh -> 0 (f(6 â€“ h) â€“ f(6)) / -h

= limh -> 0 ([6 â€“ h] â€“ [6]) / -h

= limh -> 0 (5 â€“ 6) / -h

Since [5 â€“ h] =6

= -1 / -0

= âˆž

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

Problem 2: f(x) =  Show that the above function is not derivable at x = 0.

Solution:

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

Rf'(0) = limh -> 0 f(0 + h) â€“ f(0)

= limh -> 0 (f(h) â€“ 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 letâ€™s find out the Lf'(0) and then we will check whether the function is differentiable or not.

Lf'(0) = limh -> 0 { f(0 â€“ h) â€“ f(0) } / -h

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

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

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

= (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) â‰  Lf'(0) (-1â‰ 1). Since Rf'(0) â‰  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 â€“ x of x â‰¥ 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 â€“ x). Letâ€™s find out Lf'(2) and Rf'(2)

Lf'(3) = limh -> 0 {f(3 â€“ h) â€“ f(3)} / -h

= limh -> 0 [[(2 â€“ h) + 1] â€“ [5 â€“ 2]] / -h

= limh -> 0 (3 â€“ h â€“ 3) / -h

= limh -> 0 -h / h

= -1

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

= limh -> 0 [[5 â€“ (3 + h)] â€“ 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 â€“ 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 Rfâ€™, on solving the Rf'(3) we get the value 1. Since, Rf'(3) â‰  Lf'(3) so we can say the function f(x) is not differentiable at x = 3.