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

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To prove that the imaginary part of ln(-|x|) equals π for all real x, one must recognize that negative real numbers can be expressed in polar form as re^(iπ). Direct calculation is problematic because it overlooks the multi-valued nature of the logarithm in the complex plane. By taking the logarithm, ln(r) + iπ emerges, confirming that the imaginary part consistently equals π. This approach highlights the importance of understanding complex logarithms and their properties. Thus, the proof effectively demonstrates that Im(ln(-|x|)) = π for all real x.
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How would someone go about proving:

\Im (ln(-|x|))=\pi for all reals, x, when the answer takes the form, a + bi.
 
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What's wrong with direct calculation?
 
epkid08 said:
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.
 

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