1. Partial Fraction Expansion All the polynomials in this note are assumed to be complex polynomials.

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1 Partial Fraction Expansion All the polynomials in this note are assumed to be complex polynomials A rational function / is a quotient of two polynomials P, Q with 0 By Fundamental Theorem of algebra, = q(z a ) m (z a r ) mr, where a,, a r are distinct complex numbers m i Proposition Suppose = Q (z)q 2 (z) is a product of polynomials such that Q (z), Q 2 (z) are relatively prime, ie the greatest common divisor of Q (z), Q 2 (z) is Then there exist P (z), P 2 (z) so that = P (z) Q (z) + P 2(z) Q 2 (z) Proof Since Q (z) Q 2 (z) are relatively prime, using division algorithm, we can find u (z), u 2 (z) so that Q (z)u (z) + Q 2 (z)u 2 (z) = Then = = Q (z)u (z) + Q 2 (z)u 2 (z) Q (z)q 2 (z) = u 2(z) Q (z) = P (z) Q (z) + P 2(z) Q 2 (z), + u (z) Q 2 (z) where P (z) = u 2 (z) P 2 (z) = u (z) We complete the proof Corollary Assume = (z a ) m (z a r ) mr where a,, a r are distinct Then () = P (z) (z a ) m + + P 2(z) (z a r ) mr Proof We will prove the statement by induction on deg Q Let Q (z) = (z a ) m Q 2 (z) = (z a 2 ) m2 (z a r ) mr By Lemma, we can find P (z) P (z) so that = P (z) (z a ) m + P (z) Q 2 (z) Since deg Q 2 < deg Q, by induction hypothesis, we can write P (z) Q 2 (z) = P 2(z) (z a 2 ) m P r(z) (z a r ) mr for some polynomials P 2 (z),, P r (z) This implies () Let us study the case /(z a) m first We may assume that deg P m Using division algorithm, we can find a 0,, a m so that (2) = a m + a m (z a) + + a (z a) m Then we obtain that = a z a + a 2 (z a) a m (z a) m

2 2 In reality, how do we compute a 0,, a m? In fact, d i a i = (i )! dz i (z a)m z=a Or, we can compute these numbers a,, a m using residues First, we observe that m (z a) dz = a m where r is small Denote By (2), Hence we find C r(z) R (z) = a m (z a) m = a m (z a) m a m = a m (z a) + + a (z a) m R (z) = a m + a m 2 (z a) + + a (z a) m 2 (z a) m Denote P (z) = a m + a m 2 (z a) + + a (z a) m 2 Q (z) = (z a) m Then we have P (z) Q (z) = a z a + a 2 (z a) a m (z a) m Similarly, we can compute a m using residue By induction, we can compute all a i In general, we consider / of the form () Let f (z) be the rational function f (z) = P 2(z) (z a 2 ) m P r(z) (z a r ) mr Then we can write = a a 2 + z a (z a ) a m (z a ) m + f (z) Notice that a is not a pole of f (z) For r small enough, f(z) is holomorphic on D r (a ) For 0 i m, one has d i z a i! dz i f (z)(z a ) m = 0 Using the previous observation Similarly, for s r, one has Here we write d i z a i! dz i (z a ) m = a i d i z a s i! dz i (z a s) ms = a s i m s i= a s i (z a s ) i + f s(z), f s (z) = i s P i (z) ( ) m i Or using the residue computation, for r small enough so that D r (a i ) excludes all a j for j i Then C r(a i ) () mi dz = a i m i

3 If we set H i (z) = j i (z a j) m j, then () mi = H i (z) We know /H i (z) is holomorphic on D r (a) By Cauchy integral formula, we obtain /H i (z) dz = P (a i) H i (a i ) C r(a) This gives a i m i = P (a i )/H i (a i ) Now, subtract R(z) from a i m i /( ) m i, we obtain R (z) = Inductively, we can find all a i s for s m i a i mi m i ( ) m = a i j i ( ) j + f i(z) j= Example Find the partial fraction expansion of Let f (z) f 2 (z) be rational functions f (z) = A z + 3z2 + z + (z ) 2 (z 2) 3 B (z ) 2, f 2(z) = C z 2 + so that f (z) + f 2 (z) Multiplying R(z) by z, we find D (z 2) 2 + E (z 2) 3 R(z)(z ) = z 3z2 + z + (z 2) 3 = A + B z + f 2(z)(z ) Let r be small enough so that 2 does not belong to D r () hence f 2 (z) is holomorphic on D r () By Cauchy-integral formula, B = R(z)(z )dz = ( 2) 3 = 5 C r() Subtracting R(z) from 5/(z 2) 2, we obtain By simple calculation, 5 (z 2) 2 = A z + f 2(z) 5 (z 2) 2 = 3z2 + z + + 5(z 2) 3 (z ) 2 (z 2) 3 = 5z3 27z 2 + 6z 39 (z ) 2 (z 2) 3 = 5z2 22z + 39 (z )(z 3) 2 By Cauchy integral formula by the fact that f 2 (z) is holomorphic on D r (), A = 5z 2 22z dz = (z )(z 2) 3 ( 2) 3 = 22 On the other h, Now, we know f 2 (z) = C r() ( 5 ) (z ) 2 22 z = 5z2 22z + 39 (z )(z 2) 3 22 z f 2 (z) = 5z2 22z + 39 (z )(z 2) 3 22 z = 22z2 05z + 37 (z 2) 3 3

4 4 Denote = 22z 2 05z + 37 Note that P (2) = 5, P (2) = 7 P (2)/2! = 22 We have (Taylor expansion of at z = 2) Hence we find We conclude that f 2 (z) = 3z 2 + z + (z ) 2 (z 2) 3 = = 5 7(z 2) + 22(z 2) 2 (z 2) 3 = 22 z (z 2) (z 2) 3 ( 22 z + 5 ) ( 22 (z ) 2 + z (z 2) ) (z 2) 3 Proposition 2 (Special Case) Let a,, a r be distinct complex numbers = (z a ) (z a r ) Then r = P (a i )/Q (a i ) Proof Let us assume that i= = c + + c r z a z a r Then we know ( ) = c + + c i + c i + c i+ + + c r z a + z a n Taking it z a i, we obtain On the other h, by Q(a i ) = 0, z a i As long as we can show that Q (a i ) 0, Notice that ( ) z a i = c i = z ai Q(a i ) = Q (a i ) ( ) z a i = P (a i) Q (a i ) = (z a j ) Q (a i ) = = (a i a j ) 0 z ai j i j i We prove that c i = P (a i )/Q (a i ) In the above case, since has simple pole at a i, by residue calculus, P (a i ) Q (a i ) = Res a i = P (a i ) j i (a i a j ) Example 2 Find the partial fraction expansion of z 2 + z + (z )(z 2)(z 3)

5 Write A z + B z 2 + C z 3 Then A, B, C are residues of R(z) at, 2, 3 respectively Choose r small enough so that D r () does not contain 2, 3 D r (2) does not contain, 3 D r (3) does not contain, 2 By Cauchy integral formula, A = C r() R(z)dz = ( 2)( 3) = 3 2 B = R(z)dz = C r(2) (2 )(2 3) = 7 C = R(z)dz = C r(3) (3 )(3 2) = 3 2 Example 3 Find the partial fraction expansion of z 4 + The polynomial z 4 + has 4 distinct roots We will denote them by {z, z 2, z 3, z 4 } = {e iπ/4, e iπ/2, e i3π/4, e i3π/2 } Let us consider the partial fraction expansion of R(z) : z 4 + = A + A 2 + A 3 + A 4 z z z z 2 z z 3 z z 4 The zero z i of z 4 + is a simple pole of R(z), thus z z i A i = Res zi (R(z)) = z zi z 4 + = 4zi 3 Since zi 4 =, z3 i = /z i This gives A i = z i /4 It is very interesting to see that if D is a closed disk containing all z i, then C R(z)dz = 4 i= using Vieta s formula Here C is the boundary of D Res zi (R(z)) = z + z 2 + z 3 + z 4 4 = 0 5