Asked • 12/23/22

Euler Identity values.

e=-1,,,,where i=√(-1)

is called Euller's identity


This is the principle value of e.


What is the expression providing

the other values?

3 Answers By Expert Tutors

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Dayv O. answered • 12/24/22

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Roger R.

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If you read e^{πi} as a complex power a^z with base a = (2.71...), you can have multiple values. But to my knowledge, if the base is a positive real number, people always use the principal branch of the logarithm (-π < θ ≤ π). Besides, most people might read e^z not as a complex power but as the exponential function exp(z), which is single-valued, exp(πi) = -1, unambiguously!
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12/26/22

Dayv O.

agreed,,,,given exp(i*pi)=1+i*pi/1!+(i*pi)^2/2!+... the result is a single value which is -1.
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12/26/22

Mark M. answered • 12/24/22

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Dayv O.

a very valid answer is e^(i*pi)=-e^(2*pi*pi)=-3.74x10^10, kind of far away on the number line from -1. also e^(i*pi)= -2.68x10^-9=-e^(-2*pi*pi) which is very more close to zero than to -1. where is the pi in i?
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12/24/22

Raymond B. answered • 12/23/22

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Dayv O.

CORRECTION ON FIRST EQUATIONactually the answer is e^(i*pi)=(-1)*e^(2*pi*pi*k) where k=0,+/-1,+/-2,...,,,,,,,,,,,,,,,,e^(i*pi)=-1 only when k=0,,,,,,,,,,,,when k=1 e^(i*pi)=-e^(pi*pi),,,,,,,there are an infinite number of values for e^(i*pi). I will repost the question more clearly.
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12/24/22

Jon M.

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Dayv - was a bit more involved than I thought, so I removed my answer. Based on what you wrote, isn't for k=1 --> e^(i*pi) = - e^(pi*pi+2*pi)?
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12/24/22

Jon M.

tutor
Also, for k=0, since e^(i*pi) = -1, then that would mean that e^(i*pi)=(-1)*e^(pi*pi)*e^(2*pi*0) = (-1)*e^(pi*pi)*e^(0) = (-1)*e^(pi*pi). So that would mean that e^(pi*pi) = 1, right? Where can I find out more info on this? I haven't found much on the internet other than the principal value. I don't remember learning this.
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12/24/22

Mark M.

Looks like a wholel lot of pi in the i!
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12/24/22

Dayv O.

Jon, should be (-1)e^(pi*2*pi*k) so when k=1 the value is -e^2*pi*pi, any real number raised to complex number with non-zero i component will have an infinite number of values that are valid. I will clarify by answering. Mark, looks more like some good pi.
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12/24/22

Dayv O.

Jon, Mark, Raymond see Advanced Trigonometry by Robson/Durell page 252.
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12/24/22

Dayv O.

Raymond, you can bet on this,,,,e=(-1)^(-i/pi),,,,,,and e=(-e)^(-i*2*pi) since (-e)=(-1)(e). Merry holiday days!!! ,,,,,,hey do see our new neat identity?,,,,e=(-1)^(-i*(pi^-1)) where i itself is square root of minus one. That is e=minus one to the power [[minus the square root of minus one] times [pi to the power minus one]].
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12/24/22

Jon M.

tutor
Thanks for info -- interesting problem!
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12/26/22

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