Question

What is `ln(i^2)` ?

Answer 1

`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`

Answer 2

`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`