Instead of a vague #log#, let's use the natural logarithm so things simplify nicely.
We have a complex number
#z=lnln(color(red)(cospi+isinpi))#
for which we want to find the real part, #A# and the imaginary part,#B# respectively.
What we have inside of the composed #ln# function resembles Euler's identity, which states that
#e^(icolor(blue)x)=coscolor(blue)x+isincolor(blue)x#
In our particular case, #x=pi#, hence
#cospi+isinpi=e^(ipi)#
If we plug this in our original expression, we get a simplified form:
#lnln(cospi+isinpi)=lnlne^(ipi)=lnipi#
This can be expanded using the properties of logarithms:
#lnipi=lni+lnpi#
All that is left is finding #lni#.
We already know a relation between the base of the natural logarithm, #e#, and complex numbers. To find out #lni#, suppose there exists #alpha# such #i=e^(ialpha)#.
If we write these in their complex number form we can see the similarity:
#i=color(red)0+color(blue)1*i#
#e^(ialpha)=color(red)(cosalpha)+icolor(blue)(sinalpha)#
#=>{(cosalpha=0),(sinalpha=1) :}#
One solution to this would be #alpha=pi"/"2#, and since the trigonometric functions are periodic with period #rho=2npi# for any integer #n#, the set of solutions is
#alpha=pi/2+2npi#
Our complex number can be yet again written as
#z=lni+lnpi=lne^(ialpha)+lnpi=lnpi+alphai#
#:. z=color(blue)lnpi+icolor(blue)((pi/2+2npi))#, #n in ZZ#
We have solved the question in discussion:
#{(A=lnpi),(B=pi"/"2+2npi) :}#
