Complex3 Derivative

Derivatives: Let f(z) be a fn defined on a set S and S contains a nbd of z0. Then derivative of f(z) at z0, written as

f ′( z0 ),

is defined by the equation

lim f ( z ) − f ( z0 ) f ′( z0 ) = , z → z0 z − z0 provided the limit on RHS exists.

The function f(z) is said to be differentiable at z 0 if its derivative at z 0 exists. If z - z 0 = ∆z , then (1) reduces to lim f ( z0 + ∆z ) − f ( z0 ) f ′(z 0 ) = ∆z → 0 ∆z

Qs. Differentiablity ⇒ Continuity Continuity ⇒ / Differentiablity Proof : Let f(z) is differentiable at z 0 lim f ( z ) − f ( z0 ) ⇒ f ′( z0 ) = z → z0 z − z0

lim [ f ( z ) − f ( z0 )] Now z → z0 lim  f ( z ) − f ( z0 )  = × ( z − z0 )   z → z0  z − z0 

lim  f ( z ) − f ( z0 )  =   z → z0  z − z0   lim  × ( z − z0 )  z → z 0   = f ′( z 0 ) × 0 = 0

lim ⇒ f ( z ) = f ( z0 ) z → z0 ⇒ f ( z ) is continuous at z0

Continuity

⇒ / Differentiability

Consider the function f(z) = z2 = x2+y2 = u(x,y)+ i v(x,y)

2

2

⇒ u ( x, y ) = x + y , v( x, y ) = 0. Since u and v are continuous everywhere, hence f(z) is continuous everywhere

For z ≠ z 0 , we have 2

f(z) − f(z 0 ) z − z0 = z − z0 z − z0

2

z z − z0 z0 = z − z0

z z − z z0 + z z0 − z0 z0 = z − z0 z ( z − z0 ) + z0 ( z − z0 ) = z − z0

∆z = z + z0 . , z − z0 = ∆z ∆z ∆x − i∆y = z + z0 . ∆x + i∆y y

∆

∆z plane

c1

lim f ( z ) − f ( z0 ) z → z0 z − z0 f ( z0 + ∆ z ) − f ( z0 ) = ∆z → 0 ∆z lim

 z + z0 along the path C1 = z − z only the path C  0 2

Thus, if z 0 ≠ (0,0), then lim f ( z ) − f ( z0 ) z → z0 z − z0 is not unique.

When z 0 = (0,0), then lim f ( z ) − f ( z0 ) = z 0 = 0. z → z0 z − z0 ⇒ f ( z ) is differentiable at the origin and no where else.

Sec 19 : Differentiation Formula d 1. f ′(z) = f ( z) dz d ( c ) = 0, 2. dz d ( z ) = 1, 3. dz

( )

4.

d n n −1 z = nz , dz

5.

d d ( cf ( z ) ) = c f ( z ) dz dz

d ( f(z) ± g(z)) = f ′( z ) ± g ′( z ) 6. dz d ( f ( z ) g ( z ) ) = f ( z ) g ′( z ) + f ′( z ) g ( z ) 7. dz

d  f ( z)    8. dz  g ( z ) 

g ( z ) f ′( z ) − g ′( z ) f ( z ) = , 2 ( g ( z )) if g ( z ) ≠ 0

Chain Rule: Let F(z) = g(f(z)), and assume that f(z) is differentiable at z0 & g is differentiable at f(z0), then F(z) is differentiable at z0 and

F ′( z0 ) = g ′( f ( z0 )) f ′( z0 )

Ex. Let w = f(z) and W = g(w) ⇒ W = F ( z ), hence by Chain rule dW dW dw = dz dw dz

Q 8 : (a ) f ( z ) = z , shows that f ′(z) does not exist at any point z.

Solution : Let z ≠ z 0 , then f ( z ) − f ( z0 ) z − z0 z − z0 = = z − z0 z − z0 z − z0

f ( z0 + ∆ z ) − f ( z0 ) ∆ z ⇒ = ∆z ∆z ∆ x − i∆ y = ∆ x + i∆ y

∆y

∆z = ∆x+ ∆y C2 C1

∆x

lim f ( z0 + ∆z ) − f ( z0 ) ∴ ∆z → 0 ∆z  1 along C1 = − 1 along C2 ⇒ f ′( z ) does not exist any where

Q.9 Let f be a function defined by  ( z) , z≠0  z  f ( z) =   0, z=0   Show that f ′(0) does NOT exist. 2

We have , lim f (0 + ∆z ) − f (0) f ′(0) = ∆z → 0 ∆z 2

lim (∆z ) / ∆z = ∆z → 0 ∆z

2

lim (∆ z ) ⇒ f ′(0) = 2 ∆ z → 0 (∆ z ) lim ( ∆ x − i∆ y ) = 2 (∆ x, ∆ y ) → (0,0) ( ∆ x + i∆ y ) 2

 1, along real axix  ⇒ f ′(0) =  1, along Im. axix  − 1, along line ∆ y = ∆ x  Hence f ′(0) does NOT exist.