Please, solve this for me: \ln (-1) = i\pi; and \ln (-x)

korporasidn

korporasidn

Answered question

2021-11-20

Please, solve this for me:
ln(1)=iπ;
and ln(x)=lnx+iπ for x>0 and xR

Answer & Explanation

Befory

Befory

Beginner2021-11-21Added 19 answers

Use this formula: eiπ=1
So ln(1)=iπ is reasonable. Note that ln(1)=i(2k+1)π for some kZ is just as reasonable.
For the second part you can enforce ln(ab)=ln(a)+ln(b) where a = −1 and b = x.
Michele Grimsley

Michele Grimsley

Beginner2021-11-22Added 19 answers

Try this:
eln(1)=eπi
1=eπi
eπi+1=0

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