 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-1/1c/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Log i = iπ /2

, and set

a = \frac{2 l o g 2}{π}, z = - 1 - \frac{2 i}{π} lo g a .

i^{z} = e^{z Log i} = e^{l o g a - iπ /2} = - ia,

and therefore

i^{, i^{z}} = e^{(- ia) iπ /2} = e^{aπ /2} = e^{l o g 2} = 2.

Other choices of branch of the complex logarithm produce further valid answers.

 New to topics? Read the docs here!