Question

Please bear with me. Thank you.

Evaluate:

Z=(ln(j3) / (2+j))^ 1/j


Only j3 is inside of ln function. Then it is divided by 2+j . Then the whole equation is raised to the power of 1/j

Answer 1

wolframalpha shows many ways of writing the result:

http://www.wolframalpha.com/input/?i=%28ln%283i%29%2F%282%2Bj%29%29^%281%2Fi%29

Using i instead of j,

ln(3i) = ln3 + π/2 i = 1.917cis(0.96)
2+i = 2.236cis(0.46)

ln(3i)/(2+i) = 0.857 cis(0.50)

1/i = -i, so that is

1/0.857^i cis(-0.50 i)
= (0.988 + 0.154i) * 1.649
= 1.629 + 0.254i