Kaiya Hardin

2022-04-28

Is it standard to say $-i\mathrm{log}(-1)$ is $\pi$ ?

prstenast4yi

Beginner2022-04-29Added 12 answers

It is not which $-1$ but which log which is the issue. Wolfram Alpha has used some kind of principal value of the complex logarithm. But using this, $a\mathrm{log}\left(b\right)=\mathrm{log}\left({b}^{a}\right)$ is no longer always true.

Deandre Barron

Beginner2022-04-30Added 17 answers

The (principal value) of the complex logarithm is defined as

$\mathrm{log}z=\mathrm{ln}\left|z\right|+iArg\left(z\right).$

Therefore,

$\mathrm{log}(-1)=\mathrm{ln}|-1|+iArg(-1)=0+i\pi .$

and then, one simply gets

$-i\mathrm{log}(-1)=-i\left(i\pi \right)=\pi .$

Therefore,

and then, one simply gets

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