PUMaC 2011 · 代数（B 组） · 第 4 题

PUMaC 2011 — Algebra (Division B) — Problem 4

[ 4 ] Let f be an invertible function defined on the complex numbers such that 2 z = f ( z + f ( iz + f ( − z + f ( − iz + f ( z + . . . ))))) for all complex numbers z . Suppose z 6 = 0 satisfies f ( z ) = z . Find 1 /z . (Note: an invertible 0 0 0 0 function is one that has an inverse).

解析

Substituting iz in the equation gives 2 − z = f ( iz + f ( − z + f ( − iz + f ( z + f ( iz + . . . ))))) . We then have 2 2 f ( z − z ) = f ( z + f ( iz + f ( − z + f ( − iz + f ( z + . . . = z 2 2 for all complex z . In particular, there exists some z such that z = z . We see that f ( z − z ) = 0 2 2 − 1 2 2 2 z = f ( z ). But f is one-to-one, so applying f to both sides, z − z = z . Thus, z = 2 z , 2 from which we get z = 1 / 2. Thus, 1 /z = 1 /z = 4 . 0 1 k 6 − k