Using the laws of logarithms,
#log(-1+i)-(-1-i)=log((-1+i)/(-1-i))#
Amplify the complex number by its conjugate:
#(-1+i)/(-1-i) * color(red)((-1+i)/(-1+i))=(-1+i)^2/(1^2-i^2)=(1-2i-1)/(1+1)=-(2i)/2=-i#
#:. log((-1+i)/(-1-i)) = log(-i)#
The base of the logarithm hasn't been stated as of yet, but I assume it is the natural logarithm. From now onwards, I will use #ln# to denote it.
We want to find:
#ln(-i)#
We know that the Polar form of a complex number is
#z=r(costheta+isintheta)=re^(itheta)#
#=> lnz = lnr+itheta#
Let #z=-i#.
#ln(-i) = lnr+itheta=ln|-i|+itheta=itheta#
Returning back to our trigonometric form:
#-i = costheta+isintheta#
#color(red)0color(blue)(-1)i =color(red)costheta+icolor(blue)sintheta#
#=> {(costheta=0),(sintheta=-1) :}#
The principal value of #theta# is the smallest positive one, hence:
#theta = (3pi)/2#
is the principal value.
As such, the general value of #theta# is:
#theta=(3pi)/2 +2kpi#, #kin ZZ#.
Therefore:
#ln(-i) = ln(-1+i)-ln(-1-i) = itheta=i((3pi)/2+2kpi)#
#"Principal value: " ln(-1+i)-ln(-1-i)=i(3pi)/2#
#"General value: " ln(-1+i)-ln(-1-i) =i((3pi)/2+2kpi)#
