Complex analysis
- 1. Complex Analysis ( 15UMTC61) -K.Anitha M.Sc., M.Phil., -A.Sujatha M.Sc., M.Phil., Department of Mathematics ( SF ) V.V.Vanniaperumal College for Women, Virudhunagar.
- 2. Limits • Let f be a function defined at all points z in some neighbourhood of 𝑧0 except possibly for the point 𝑧0itself. • The limit of f(z) as z approaches 𝑧0is a number 𝑤0, • ie, lim 𝑧→𝑧0 𝑓 𝑧 = 𝑤0
- 3. Uniqueness of limit • When limit of a function f ( z ) exists at a point 𝑧0 , it must be unique. Proof: Suppose f ( z ) has two different limits 𝑤0 and 𝑤1 Then for a positive number 𝜀 , there are + ve number’s 𝛿0 and 𝛿1 such that 𝑓 𝑧 − 𝑤0 < 𝜀 whenever 0 < 𝑧 − 𝑧0 < 𝛿0 And 𝑓 𝑧 − 𝑤1 < 𝜀 whenever 0 < 𝑧 − 𝑧0 < 𝛿1
- 4. Let 𝛿 = min (𝛿0 , 𝛿1) . Then for 0 < 𝑧 − 𝑧0 < 𝛿0 • 𝑓 𝑧 − 𝑤0 − 𝑓 𝑧 − 𝑤1 ≤ 𝑓 𝑧 − 𝑤0 + 𝑓 𝑧 − 𝑤1 < 2𝜀 ie, 𝑤1 − 𝑤0 < 2𝜀 But 𝑤1 − 𝑤0 is a constant and 𝜀 can be chosen arbitrarily small. If we choose 𝜀 = 0 , then 𝑤1 − 𝑤0= 0 ie , 𝑤1 = 𝑤0 Hence the limit of a function is unique.
- 5. Continuity • The function f ( z ) is continuous at a point 𝑧0 , if i) lim 𝑧→𝑧0 𝑓(𝑧) exists ii) f(𝑧0) exists iii) lim 𝑧→𝑧0 𝑓(𝑧) = f(𝑧0) A function of a complex variable is said to be continuous in a region R if it is continuous at each point in R.
- 6. Properties i) Sum of two continuous functions is a continuous function . ii) Product of the continuous functions is a continuous function . iii) Quotient of two continuous functions is continuous where the denominator not equal to zero. iv) Polynomial is continuous in the entire plane.
- 7. Derivatives • The derivative of f at 𝑧0, denoted by 𝑓/(𝑧0) = lim 𝑧→𝑧0 𝑓 𝑧 −𝑓(𝑧0) 𝑧−𝑧0 provided this limit exists. • The function f is said to be differentiable at 𝑧0 when its derivative at 𝑧0 exists.
- 8. Analytic Functions • A single valued function w = f ( z ) in a domain D is said to be analytic at a point z = a in a D if there exists a neighbourhood 𝑧 − 𝑎 < 𝛿 at all points of which the function is differentiable. Singular Point The point at which f ( z ) is not differentiable are called singular points of the function.
- 9. Ex. Given u = 𝑦3 - 3𝑥2 y find f(z)such that f(z) is analytic • Solution • dv = 𝜕𝑣 𝜕𝑥 𝑑𝑥 + 𝜕𝑣 𝜕𝑦 dy = −𝜕𝑢 𝜕𝑦 𝑑𝑥 + 𝜕𝑢 𝜕𝑥 dy ( C R equations) v = ( −𝜕𝑢 𝜕𝑦 𝑑𝑥 + 𝜕𝑢 𝜕𝑥 dy ) 𝜕𝑢 𝜕𝑥 = -6xy ; 𝜕𝑢 𝜕𝑦 = 3𝑦2 -3𝑥2 v = (−3𝑦2+3𝑥2)𝑑𝑥 − 6𝑥𝑦 𝑑𝑦
- 10. = 3𝑥2 dx – (3𝑦2 dx + 6xy dy ) = 𝑑 ( 𝑥3) - d ( 3x𝑦2 ) v = 𝑥3 - 3x𝑦2 + c Hence f ( z ) = u + iv = 𝑦3- 3𝑥2y + i (𝑥3 - 3x𝑦2) _____________________