By R. Vichnevetsky, et al.,

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Let n be a positive integer. Near a we can write k! k! Let ck = ❺ ❺✑❺✝❺✝❺✝❺ ∞ n g( z ) = P( z ) + r( z ), P( z ) = ∑ c k ( z − a) k, r( z ) = k =1 ∑ ck (z − a)k, k =n +1 noting that c0 = g(a) = 0. Near b = 0 = g(a) we can write n f (w) = Q(w) + s(w), Q(w) = ∑ dk w k, s(w) = k =0 ∞ ∑ dk w k. k =n +1 Here P and Q are polynomials and r is analytic near a , while s is analytic near 0. For z close to a we know that g( z ) is close to 0. If we substitute w = g( z ) then for z close to a we have h( z ) = f (g( z )) = ( ) n ∑ dk (P(z) + r(z))k + s(g(z)) = R(z) + s(g(z)).

Near b = 0 = g(a) we can write n f (w) = Q(w) + s(w), Q(w) = ∑ dk w k, s(w) = k =0 ∞ ∑ dk w k. k =n +1 Here P and Q are polynomials and r is analytic near a , while s is analytic near 0. For z close to a we know that g( z ) is close to 0. If we substitute w = g( z ) then for z close to a we have h( z ) = f (g( z )) = ( ) n ∑ dk (P(z) + r(z))k + s(g(z)) = R(z) + s(g(z)). k =0 Now s(0) = s ′ (0) = .... = s (n)(0), because s is a power series with first term dn + 1 w n + 1. We calculate the p ’th derivative of s(g) at a , where 1 p n , using the Claim above, with F = s and G = g and m = p .

Consider ∫ 4. Consider ∫ 5. Consider ∫ z − 17 dz . ( z −2)( z − 4) ➧➨➧☞➧✑➧✝➧✑➧✝➧✝➧✝➧✑➧ ➥ z ➦ = 300 1 dz . e −1 ➭ ➭☞➭✑➭✝➭ z ➩ z➫ = 1 1 dz . (e − 1)2 ➳➵➳☞➳✝➳✑➳✝➳✑➳ z ➯ z➲ = 1 6. Let γ be the semicircular contour through − R,R and iR, and calculate 7. Determine lim ∫ R R → ∞ 0 8. Evaluate ∫ ∞ −∞ (cos x) / (x 2 + 1) dx . 1 / (x 2 + 2x + 6) dx . ∫ γ e iz / ( z 2 + 1)dz .