This document explains the concept of Laurent series expansion in complex analysis. It covers the conditions for convergence and the radius of convergence. Solved examples and practice problems are provided for better understanding.
Contribute Materials
Your contribution can guide someone’s learning journey. Share your
documents today.
1)b) d dzlnz=1 z=1 a+(z−a)=1 a∑(a−z a) n a=1+2i lnz=∫1 a∑(a−z a) n dz=¿∑1 n(a−z a) n +lna¿ Radius of convergence: Ratio test or root test will both give you the same thing. |a−z|<|a|=¿R<|a|=¿R<√5 a) 1 z−a−(1−a)=1 (a−1)∑(z−a a−1)n a=1+2i For convergence, we can say that, ¿z−a∨¿ ¿a−1∨¿<1¿¿ R<|a−1|=|1+2i−1|=|2i|=2=¿R<2 c) z2−1 z3−1=z+1 z2+z+1 About z = 2 Therefore,z=w+2 w+3 w2+5w+7=w+3 (w+5+i√3 2)(w+5−i√3 2) ¿A 5+i√3 2 ∑(w 5+i√3 2)n +B 5−i√3 2 ∑(w 5−i√3 2)n |(w 5+i√3 2)|<1
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
|w|<|5±i√3 2|=√7 Therefore the radius of convergence is√7 2) 1 z−3−z z−3 Letz=w+1 1 w−2−w+1 w−2=−w w−2 ¿w 2(1 1−w 2) ¿w 2∑ n=0 ∞ (w 2)n ¿∑ p=0 ∞ (w 2)p wherep=n+1 |w 2|<1=¿|w|<2 Radius of convergence is|z−1|<2 3)a) lnz=∫1 a∑(a−z a) n dz=¿∑1 n(a−z a) n +lna¿ a=i−2 b) |(a−z a)|<1 |z−a|<|a|=|i−2|=√5 c) If a = -2,
Thenlnadoesnot∃. 4) f'(x)=lim h→0 f(x+h)−f(x) h f'(x)=lim h→0 e −1 (x+h)2 −e −1 (x)2 h Expanding, e −1 (x+h)2 =1−1 (x+h)2+1 (x+h)4−… e −1 (x)2 =1−1 (x)2+1 (x)4−… e −1 (x+h)2 −e −1 (x)2 =(2x+h)h (x+h)2(x)2+…somehigherordertermsofh f'(x)=lim h→0 (2x+h)h (x+h)2(x)2+…somehigherordertermsofh h f'(x)=(2x) (x)2(x)2(1−1 (x)2+1 (x)4−…) f'(x)=2 x3(1−1 (x)2+1 (x)4−…)=2 x3e −1 (x)2 Therefore f'(x)=2 x3e −1 (x)2 At x = 0, putting x = 0 inf'(x), e −1 (x)2 →0asx→0 So we can write it as, f'(0)=2 x3e −1 (x)2 =0
b) With the help of induction we can show that higher order derivatives are also zero so, f'(0)=f''(0)=f'''(0)=…=0 Now if we find the Taylor series off(x)at 0, f(0+h)=f(0)+hf'(0)+h2 2f''(0)+h3 6f'''(0)+… Putting the values of the derivatives we get, f(0+h)=0 c) lim h→0 f(0+h)−f(0)=lim h→0 e −1 h2 As h is imaginary soh2will give a negative value so it will become -h2 lim h→0 e 1 h2 =∞≠f(0) Therefore it is not continuous proved. d) e −1 z2 =1+−1 z2+1 2!z4+−1 3!z6+… ¿∑ n=0 ∞(−1)n n!(z2)n=∑ n=0 ∞1 n!(−1 z2) n It is valid in the domain of |z| > 0 5) 1 z2−5z+6=1 (z−3)(z−2)=1 z−3−1 z−2 If |Z|> 3 1 z−3=1 z(1 1−3z−1)=1 z∑ n=0 ∞ (3 z)n =∑ n=0 ∞3n zn+1
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser