Question: Show that (a) log e = 1 + 2ni (n = 0,1,2, . . .); (b) log I = (2n + ) i (n =
(a) log e = 1 + 2nπi (n = 0,±1,±2, . . .);
(b) log I = (2n + ½) πi (n = 0, ±1, ±2, ...);
(c) log (- 1 + √3i)} = ln 2 + 2(n + 1/3) πi (n = 0, ±1, ±2, ...).
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