Interesting question: why is ln(-1) in polar form....?

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The discussion centers on the mathematical concept of the natural logarithm of negative one, specifically ln(-1), which equals πi. Participants highlight the relationship between complex numbers and polar coordinates, noting that ln(-1) can be expressed in polar form as πe^(iπ/2). This connection illustrates the fundamental intertwining of exponential functions and complex numbers in mathematics.

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rabualeez
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Hi all,
I was doing some math and I stumbled upon a very interesting thing. When I do ln(-1), I get πi, and when I turn that into polar coordinates on the calculator, it gives me πeiπ/2 . Why is that? I'm very curious to know, because they are so intertwined!

Thank you
 
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Hello Rabu, :welcome:

Apparently, ##i## and ##e^{i\pi/2}## are one and the same. Can you see why ?
The calculator responds in a form ##|\alpha| \,e^{\operatorname{\, Arg}(\alpha)}## which I think is reasonable for complex numbers.
 
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