I had a question about a way to write Euler's identity

  • #1
jeremyrijsdijk
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since euler's identity states e^pi*i+-1 is it okay to write down e^pi*= e * -1/e = -1 and would for integers the same rule apply where a^pi*i= a * -1/a = -1 or is it only for the constant e?
this might be a stupid question, but i cant find an answer and im curious
 
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  • #2
jeremyrijsdijk said:
since euler's identity states e^pi*i+-1 is it okay to write down e * -1/e = -1
That is not right at all! You can evaluate that expression on a calculator.

jeremyrijsdijk said:
would for integers the same rule apply where a^pi*i= a * -1/a = -1
Definitely not!
 
  • #3
The correct identity using Latex to display it properly:

##e^{i\pi} + 1 = 0##

It contains five of the most famous mathematical constants:

- 0 The additive identity
- 1 The multiplicative identity
- i The imaginary unit
- ##e## base of the natural logarithms
- ##\pi## ratio of the circle circumference to its diameter

What's remarkable is that ##e## and ##\pi## are both transcendental numbers that, when combined above, produce an integer value.
 
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  • #4
jeremyrijsdijk said:
e^pi*= e * -1
PeroK said:
That is not right at all! You can evaluate that expression on a calculator.


Definitely not!
for any integer a * - 1/a = -1 so why isnt e^pi*i the same as the formula a *-1/a? since it has the same outcome? im not a math major, im in highschool i was just curious.
 
  • #5
jeremyrijsdijk said:
im not a math major, im in highschool
(I Changed the thread prefix from "A" = Advanced/Graduate School level discussion to "B" = Basic High School level discussion.)
 
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  • #6
PeroK said:
That is not right at all! You can evaluate that expression on a calculator.
It is correct (when the parentheses OP should have introduced are placed and the missing i is returned). However, the question is why OP would think it is even relevant to rewrite it like that …
 
  • #7
Orodruin said:
It is correct (when the parentheses OP should have introduced are placed and the missing i is returned). However, the question is why OP would think it is even relevant to rewrite it like that …
I was wondering if there is a way to get a positive integer to be negative after purely powering it by a certain value, and since euler’s identity is the only thing I could find which gave a negative result after a power, I wondered if imaginary numbers are able to have such qualities. I don’t get all the scrutiny, I think it’s unnecessary for such an innocent question.
 
  • #8
jeremyrijsdijk said:
I don’t get all the scrutiny, I think it’s unnecessary for such an innocent question.
Not really "scrutiny", just a bunch of people with advanced degrees (check our Profile/About pages) trying to help a newbie with their inquisitive questions. I'll send you a PM to help you understand how to use LaTeX to post math here at PF, which will improve your questions a lot. :smile:
 
  • #9
jeremyrijsdijk said:
I was wondering if there is a way to get a positive integer to be negative after purely powering it by a certain value, and since euler’s identity is the only thing I could find which gave a negative result after a power, I wondered if imaginary numbers are able to have such qualities. I don’t get all the scrutiny, I think it’s unnecessary for such an innocent question.
You can read about complex exponents here:

https://brilliant.org/wiki/complex-exponentiation/
 
  • #10
berkeman said:
Not really "scrutiny", just a bunch of people with advanced degrees (check our Profile/About pages) trying to help a newbie with their inquisitive questions. I'll send you a PM to help you understand how to use LaTeX to post math here at PF, which will improve your questions a lot. :smile:
The question was just if a^iπ = a -\left( \frac 1 a\right)? Is there someone who can explain it to me? LaTeX does not seem to work from the device I am using, therefore I can not improve my questions your majesty. I was just looking for an answer to a simple question.:thumbup:
 
  • #11
jeremyrijsdijk said:
The question was just if a^iπ = a -\left( \frac 1 a\right)? Is there someone who can explain it to me? LaTeX does not seem to work from the device I am using
You forgot to put pairs of # or $ marks before and after. Adding them, we get ##a^iπ = a -\left( \frac 1 a\right)## (quote my post to see the LaTeX for that). I suspect you actually meant ##a^{iπ}= a \left( \frac {-1} a\right)##, which is a long winded way of writing ##a^{i\pi}=-1##, which is not correct, no.

Generally, you can write a complex number as ##Ae^{i\theta}## where ##A## is a non-negative real number and ##\theta## is an angle in radians. So we can put your ##a^{i\pi}##into a more convenient form by finding ##A## and ##\theta## for this case: $$\begin{eqnarray*}
a^{i\pi}&=&Ae^{i\theta}\\
i\pi\ln(a)&=&i\theta+\ln(A)
\end{eqnarray*}$$where we get the second line by taking the natural log of the first. Comparing real parts in the second line we get ##0=\ln(A)## which means ##A=1##. Comparing imaginary parts we get ##i\pi\ln(a)=i\theta## which means ##\theta=\pi\ln(a)##. So generally, ##a^{i\pi}=e^{i\pi\ln(a)}##. From the right hand side we can see that this will only be ##-1## if ##\ln(a)## is an odd integer - i.e. if ##a=e^n## where ##n## is odd.
 
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