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

[jeremyrijsdijk]

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

 

[PeroK]

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.

would for integers the same rule apply where a^pi*i= a * -1/a = -1

Definitely not!

 

[jedishrfu]

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.

 

[jeremyrijsdijk]

e^pi*= e * -1

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.

 

[berkeman]

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.)

 

[Orodruin]

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 …

 

[jeremyrijsdijk]

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.

 

[berkeman]

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.

 

[PeroK]

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/

 

[jeremyrijsdijk]

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.

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.

 

[Ibix]

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.