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 a^{2}+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 = z
^{3} - W = x
^{2}-y^{2}+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)=x
^{2}+y^{2} ^{ }G(w,t)=5wt

### Differentiable complex Function

A complex function F[z] is said to be differentiable at z=z_{0} if the following limit exists at z=z_{0}.

**lim** _{Î´zâ‡¢0}**(F(z+Î´z)-F(z))/Î´z**

exists at z=z_{0}

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)=z ^{2}.**

**Solution:**

We know that for a function to be differentiable at any z

_{0}the following limit should exist at z=z_{0 }lim

_{Î´zâ‡¢0}(F(z+Î´z)-F(z))/Î´z exists at z=z_{0}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}-z^{2}=z^{2}+2zÎ´z+Î´z^{2}-z^{2}= 2zÎ´z+ Î´z^{2}lim

_{Î´zâ‡¢0}(F(z+Î´z)-F(z))/Î´zâ‡’lim

_{Î´zâ‡¢0}(2zÎ´z+ Î´z2)/Î´zWe 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=0lim

_{ Î´xâ‡¢0}(Î´x +(2x+2iy) = 2x+2iy =2zConsider second path along vertical line(parallel to y axis) i.e., Î´ x=0.

lim

_{Î´yâ‡¢0}(Î´x +(2x+2iy) = 2x+2iy =2zSo 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)=z^{2}is differentiable at all z âˆˆ complex number.

**Note:** If F(z)=z^{n} 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)/g
^{2}

**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:

- Partial derivatives of U(x,y) and V(x,y) w.r.t x and y must exist and be continuous.
- Partial derivatives of U(x,y) w.r.t x = Partial derivatives of V(x,y) w.r.t y
- 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 z _{0} 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:

- (Î´U/Î´x) = (Î´V/Î´y)
- (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 z_{0} the given limit should exist

lim _{Î´zâ‡¢0} (F(z+Î´z)-F(z))/Î´z exists at z=z_{0}

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/(i^{2}) =-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 = z_{0} if it is differentiable at z = z_{0} and also in some neighborhoods of z_{0}.*Neighborhood: The neighborhood of a point z _{0} is a set of all points z where |z âˆ’ z_{0}| < 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) = z

^{3}+ 3z

^{2}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) = e ^{x}(cosy + i siny)**

**Solution:**

We have U (x, y) = e

^{x}cosy and V (x, y) = e^{x}siny

Î´U/Î´x = e^{x}cosy

Î´U/Î´y = âˆ’e^{x}siny

Î´V/Î´x = e^{x}siny

Î´V/Î´y = e^{x}cosy

So we can clearly see that cauchy reimann equations are satisfied

Î´U/Î´x = Î´V/Î´y = e^{x}cosy

Î´U/Î´y = âˆ’ Î´V/Î´x = e^{x}sinyAnd 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:**

- 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.
- 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)=3x^{2}y âˆ’ y^{3} + x^{2} âˆ’ 3x + i(âˆ’x^{3} + 3xy^{2} + 2y^{2}) differentiable? Where is it analytic?

**Answer:**

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

F (z) = 3x^{2}y âˆ’ y^{3}+ x^{2}âˆ’ 3x + i(âˆ’x^{3}+ 3xy^{2}+ 2y^{2})

U (x, y) = 3x^{2}y âˆ’ y^{3}+ x^{2}âˆ’ 3x and V (x, y) = âˆ’x^{3}+ 3xy^{2}+ 2y^{2}

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 = 3y^{2}âˆ’ 3x^{2}

Partial derivatives of U(x,y) w.r.t y= Î´U/Î´y = 3x^{2}âˆ’ 3y^{2}

Partial derivatives of V(x,y) w.r.t y= Î´V/Î´y = 6xy + 4yPartial 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 = 3x^{2}âˆ’ 3y^{2}

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 line2x âˆ’ 3 = 4y, not through any region (neighborhood)R. Hence F (z) isnowhereanalytic.

### 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€™ = lim

_{h -> 0}f(5 + h) â€“ f(5)= lim

_{h -> 0}[5 + h] â€“ [5]Since [h + 5] = 5

= lim

_{h -> 0}(5 â€“ 5) / h = 0Lf'(1) = lim

_{h -> 0}[f(5 â€“ h) â€“ f(5)] / -h= lim

_{h -> 0}( [5 â€“ h] â€“ [5] ) / -hSince [5 â€“ h] = 4

= lim

_{h -> 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) = lim

_{h -> 0}f(6 + h) â€“ f(6)= lim

_{h -> 0}([6 + h] â€“ [6]) / hSince, 6 + h = 6

= lim

_{h -> 0}(6 – 6) / h= lim

_{h ->0}0 / h=0

Lf'(6) = lim

_{h -> 0}(f(6 â€“ h) â€“ f(6)) / -h= lim

_{h -> 0 }([6 â€“ h] â€“ [6]) / -h= lim

_{h -> 0}(5 â€“ 6) / -hSince [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) = lim

_{h -> 0}f(0 + h) â€“ f(0)= lim

_{h -> 0}(f(h) â€“ f(0)) / h= lim

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

_{h -> 0}(e(1 / h) + 1) / (e(1 / h) – 1)Multiply by e(1 / h)

= lim

_{h -> 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) = lim

_{h -> 0}{ f(0 â€“ h) â€“ f(0) } / -h= lim

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

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

_{h -> 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) = lim

_{h -> 0}{f(3 â€“ h) â€“ f(3)} / -h= lim

_{h -> 0}[[(2 â€“ h) + 1] â€“ [5 â€“ 2]] / -h= lim

_{h -> 0}(3 â€“ h â€“ 3) / -h= lim

_{h -> 0}-h / h= -1

Rf'(3) = lim

_{h -> 0}{f(3 + h) â€“ f(3)} / h= lim

_{h -> 0}[[5 â€“ (3 + h)] â€“ 2] / h=lim

_{h -> 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.