 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/2/a/solution

Log i = iπ /2

and

Log (- e) = 1 + iπ

. Thus the values and their real parts and imaginary parts are

expression i ln (- e) 2^{i} (i^{i})^{i} principal value (1 + i) / 2 1 + iπ e^{i l n 2} - i Re 1/ 2 1 cos (ln 2) 0 Im 1/ 2 π sin (ln 2) - 1

For the final row, first calculate

i^{i} = e^{- π /2}

and only then raise this positive real number to the power

i

. Treating nested powers as ordinary real powers would ignore the branching of complex exponentiation.

If all branches are requested,

i

also has the negative of its principal value, and the complex logarithm of

- e

1 + i (π + 2 πk)

. Allowing nonprincipal logarithms of the positive base

2

gives

e^{- 2 πk} e^{i l n 2}

. Finally, all values of

(i^{i})^{i}

are

- i e^{- 2 πm}

k, m

as appropriate. These alternatives do not alter the principal-value table.

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