# midterm - agree at α 1,α n Indeed this would say that the...

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Math 220A Complex Analysis Solutions to Midterm Prof: Lei Ni TA: Kevin McGown Midterm, Problem 4. Suppose that Q ( z ) is a polynomial of degree n with n distinct roots α 1 ,...,α n and that P ( z ) is a polynomial with degree less than n . Show P ( z ) Q ( z ) = n X k =1 P ( α k ) Q 0 ( α k )( z - α k ) . Proof. By the Fundamental Theorem of Algebra we have Q ( z ) = β n Y k =1 ( z - α k ) , β 6 = 0 . Our goal is to prove: P ( z ) = n X k =1 P ( α k ) Q ( z ) Q 0 ( α k )( z - α k ) ( * ) Without loss of generality, we may assume that Q ( z ) is monic (i.e. β = 1) since replacing Q ( z ) by β - 1 Q ( z ) leaves the RHS of ( * ) unchanged. First observe that z - α k divides Q ( z ) for each k and hence the RHS of ( * ) is a polynomial of degree less than n . Invoking again the Fund. Thm. of Algebra, it suﬃces to prove that the LHS and RHS of (
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Unformatted text preview: * ) agree at α 1 ,...,α n . Indeed, this would say that the diﬀerence of the two sides is a polynomial of degree less than n having n roots, which would force it to be the zero polynomial. By the product rule for derivatives, we have Q ( α k ) = Y j 6 = k ( α k-α j ) and by direct calculation ± Q ( z ) ( z-α k ) ² z = α ` = δ jk . Now we evaluate each side of ( * ) at α j to obtain: RHS : n X k =1 P ( α k ) δ jk = P ( α j ) LHS : P ( α j ) Hence the RHS and LHS of ( * ) agree at each α j and the proof is complete. ± 1...
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## This note was uploaded on 10/19/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.

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