Prove arcsin x for its logarithm form

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SUMMARY

The discussion focuses on proving the logarithmic form of the arcsine function, specifically that arcsin x = -i ln(ix + √(1 - x²)). The user references the Euler's formula, sin x = (e^(ix) - e^(-ix)) / 2i, as a foundational concept. The challenge lies in understanding complex logarithms, which the user admits to lacking knowledge about. The suggestion to work backward by substituting arcsin x into the initial equation is presented as a potential solution strategy.

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  • Understanding of Euler's formula
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  • Familiarity with the properties of logarithms
  • Basic concepts of inverse trigonometric functions
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  • Review the derivation of inverse trigonometric functions
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Homework Statement


Given sin x = (eix - e-ix) / 2i,
I want to prove that

arcsin x = -i ln(ix + √1 - x2)

Homework Equations


I know about the Euler's formula and complex number. But I have never learned about complex logarithms.

The Attempt at a Solution


I try to use x = sin y. But it seems not working. How to show that relation?
 
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Why not try working it backward? Take the arcsin x and substitute it into the first equation for x.
 

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