 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/11/a/i/solution

Use the full set of complex logarithm values of

i

lo g i = i (\frac{π}{2} + 2 πk), k \in Z .

By the definition of complex exponentiation, the inner power can therefore take any value

i^{i} = exp [- (π /2 + 2 πk)]

, a positive real number. Its complex logarithm values are

- (π /2 + 2 πk) + 2 πim

, where

m \in Z

. Thus

(i^{i})^{i} = exp [- i (\frac{π}{2} + 2 πk) - 2 πm] = - i e^{- 2 πm} .

All possible real parts and imaginary parts are consequently

ℜ ((i^{i})^{i}) = 0, ℑ ((i^{i})^{i}) = - e^{- 2 πm}, m \in Z .

Choosing the principal complex logarithm at both stages gives the particular value

- i

. The independent logarithm choices are needed because the question asks for all possible values.

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