Question

解方程lnz=2-i(π/6)

Answer 1

解就是z=e^(2-iπ/6)=e²(cosπ/6-isinπ/6)=e²(√3-i)/2

Answer 2

lnz=2-i(π/6)

z = e^(2-i(π/6))
= e^2 / e^( i(π/6))
= e^2 / [cos(π/6) + isin(π/6 ]
= e^2 . [cos(π/6) - isin(π/6) ]
= e^2 . [ (√3/2) - (1/2)i ]