Question

Prove that the real part of the principle value of (1+i)^logi is e^[(-π²/8)×cos(π/4log2)]

Answer 1

Answer:

Using complex natural logarithm, i.e., logz=log(x+iy)=log(|z|ei(ϕ+2πn))=log|z|+i(ϕ+2πn)

ilog(1+i)=elogi⋅log(1+i)=e(log1+i(π/2+2πn))⋅(log2√+i(π/4+2πk))=e12log2(iπ/2+i2πn)−(π/2+2πn)(π/4+2πk)=

Final simplification:

=e−π2/8+π2k+π2/2n+4π2nkeiπ/2log2(1/2+2n)

For the main branch, i.e., k=0, n=0, the result is

e−π28eiπlog24≈0.2912ei0.544≈0.24912+0.1508i

Step-by-step explanation:

For principal value putting n=k=0 we get ilog(1+i)=2iπ4.e−π28