Question

show that
cos(log i^i)=0

Answer 1

Step-by-step explanation:

cos(log i^i) =0

log i^i = cos^-1(0),

i log i = π/2

log i =π/2i

log i = π/2 (-1)(-1)/(√-1)

log i = π/2 (-i)

e^(log i) = e^(-iπ/2)

i = e^ [-i(π/2) ]

I = ( cosπ/2 - i sin π/2)