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fe-(3+ik) g(k) = F-'[g(k)] = f(x) a. 3 + ik F-'g(k)] = f(x) b. g ( k ) = 2+i(k- 1) = g (ki) =
\fe-(3+ik) g(k) = F-'[g(k)] = f(x) a. 3 + ik F-'g(k)] = f(x) b. g ( k ) = 2+i(k- 1) = g (ki) = ik F-' [g(k)] = f(x) C. (1 + ik) (2 + ik)
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