A familiar identity

Algebra Level pending

ln (-1) = ?

ei eπ πi πie √i sin(π)

Andrew Paul by Andrew Paul , 22, USA

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1 solution

Andrew Paul
Oct 15, 2015

Euler's identity states: e^πi + 1 = 0 So: ln (-1) = πi

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You might have to be careful here. Note e ( 2 n + 1 ) π i + 1 = 0 e^{(2n+1)\pi i}+1=0 too. So ln ⁡ ( − 1 ) = ( 2 n + 1 ) π i \ln(-1)=(2n+1)\pi i ?

We have to be careful when dealing with complex logarithms .

We get around this problem by letting the argument of the z z lie only in the range ( − π , π ] (-\pi,\pi]

Isaac Buckley - 5 years, 8 months ago

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