What is ln(i^2) ?

2 Answers
Nov 24, 2017

ln(i^2)=pii

Explanation:

ln(i)=ln(sqrt(-1))=ln((-1)^(1/2))=1/2ln(-1)=1/2*pii

so,
ln(i^2)=2*ln(i)=2*1/2*pii=pii

Nov 24, 2017

ln(i^2) = ln(-1) = pii

Explanation:

The function e^x considered as a real valued function of real numbers is one to one from (-oo, oo) onto (0, oo)

Therefore the real valued logarithm is a well defined function from (0, oo) onto (-oo, oo)

The function e^z considered as a complex valued function of complex numbers is many to one from CC onto CC "\" { 0 }.

In particular, note that e^(2pii) = 1. So if e^z = c then e^(z+2npii) = c for any integer n.

Therefore the complex logarithm has multiple branches, with a principal branch from CC "\" { 0 } onto { x+yi in CC : y in (-pi, pi] } (using the normal range of Arg(z) in (-pi, pi])

The expression ln(z) denotes this principal value.

So whereas z = 7ipi is a root of e^z = -1, it is not the principal value of ln(i^2) = ln(-1).

The principal value is ln(-1) = pii

In general, we can write a formula for the principal value of the logarithm of a complex number z as:

ln z = ln abs(z) + Arg(z) i