Question

solve both questions​

Answer 1

Answer:

solved

Step-by-step explanation:

i) now i = cosπ/2 + isinπ/2 = e^iπ/2

i^i =e^i²π/2 = e^-π/2

logi^i = log e^-π/2 = -π/2

sin(logi^i ) = sin(-π/2) = -sinπ/2 = -1

ii) from the previous question ,

logi (to basei) = R = iπ/2 loge(tobasei) =  iπ/2 loge÷logi =  iπ/2 lne÷lni

R = iπ/2 ÷lni = 1 = 4m÷4n = 4m+1 ÷ 4n+1

m = n , m&n are integers