Proving $\Im (\ln(-|x|))=\pi$ for All Reals

[epkid08]

How would someone go about proving:

$$\Im (ln(-|x|))=\pi$$ for all reals, x, when the answer takes the form, a + bi.

 

[Hurkyl]

What's wrong with direct calculation?

 

[mathman]

How would someone go about proving:

$$\Im (ln(-|x|))=\pi$$ for all reals, x, when the answer takes the form, a + bi.

Negative real numbers can be expressed in polar coordinates as re^(pi)i. Take the log and you get ln(r)+(pi)i.