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Thread: Euler's Identity and i: how to solve e^(pi*i) = -1 for i

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    Euler's Identity and i: how to solve e^(pi*i) = -1 for i

    There was something interesting that I found out when looking at Euler's Identity. If e^(pi*i) = -1 and i^2 = -1 then e^(pi*i) has to equal i^2 to get the equation seen here.

    [tex]e^{\pi \cdot i} = i^{2}[/tex]


    My question is can you in this equation solve for i and if you can what would you get? Will i equal to square root of -1 or will you get something different?
    Last edited by mmm4444bot; 01-06-2018 at 08:49 PM. Reason: Inserted missing grouping symbols (required); replaced link to objectionable site with LaTex

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    Elite Member stapel's Avatar
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    Quote Originally Posted by predarsh View Post
    There was something interesting that I found out when looking at Euler's Identity. If e^(pi*i) = -1 and i^2 = -1 then e^(pi*i) has to equal i^2 to get the equation seen here.

    [tex]e^{\pi \cdot i} = i^{2}[/tex]

    My question is can you in this equation solve for i and if you can what would you get? Will i equal to square root of -1 or will you get something different?
    What did you get when you tried?

    Please be complete. Thank you!
    Last edited by mmm4444bot; 01-06-2018 at 07:44 PM. Reason: Same as above

  3. #3
    Quote Originally Posted by stapel View Post
    What did you get when you tried?

    Please be complete. Thank you!
    I'm not sure how to do logs with complex exponents which is why I asked if anyone knew how and what the answer would be.

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    Elite Member mmm4444bot's Avatar
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    https://en.wikipedia.org/wiki/Complex_logarithm

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    Quote Originally Posted by predarsh View Post
    I'm not sure how to do logs with complex exponents which is why I asked if anyone knew how and what the answer would be.
    I think what the others are trying to do is to get you to try this out yourself. You'll learn a lot more by playing with your idea and telling others what you found, than by just asking someone for an answer. There are different directions in which you might take this, which would trigger different follow-up questions. I'll be happy to discuss it with you as you get into it.

    But one thing to consider is that if you square the identity, you get an expression equal to 1; which means that anything multiplied by that expression is unchanged. This leads to some very tricky behavior when you get into complex logs in general!

  6. #6
    Quote Originally Posted by Dr.Peterson View Post
    I think what the others are trying to do is to get you to try this out yourself. You'll learn a lot more by playing with your idea and telling others what you found, than by just asking someone for an answer. There are different directions in which you might take this, which would trigger different follow-up questions. I'll be happy to discuss it with you as you get into it.

    But one thing to consider is that if you square the identity, you get an expression equal to 1; which means that anything multiplied by that expression is unchanged. This leads to some very tricky behavior when you get into complex logs in general!
    I tried taking logs of both sides and using the property rule. I got stuck at pi*i log e = 2log i and from log e^(pi*i)=log i^2.

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    Quote Originally Posted by predarsh View Post
    I tried taking logs of both sides and using the property rule. I got stuck at pi*i log e = 2log i and from log e^(pi*i)=log i^2.
    Have you read the Wikipedia article about complex logs that was suggested to you? The "tricky behavior" I mentioned is the fact, explained there, that the log, as a function of a complex variable, is multi-valued.

    Now, looking back at your original question, it looks like what you meant was that since i satisfies the equation [tex]e^{\pi \cdot i} = i^2[/tex], you had the idea that you could solve the equation [tex]e^{\pi \cdot x} = x^2[/tex] for x, and wondered if you would find other solutions besides i. (You can't solve an equation for the value of a constant, like solving 2(3) = 6 for 3, so the way you stated the question was confusing.) I'm not sure whether that equation can be solved at all; if so, it would involve some very ugly manipulations!

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    Alpha

    "What happens in the event horizon, stays in the event horizon" -- Bob Brown (grandpa Bob)

  9. #9
    Quote Originally Posted by Dr.Peterson View Post
    Have you read the Wikipedia article about complex logs that was suggested to you? The "tricky behavior" I mentioned is the fact, explained there, that the log, as a function of a complex variable, is multi-valued.

    Now, looking back at your original question, it looks like what you meant was that since i satisfies the equation [tex]e^{\pi \cdot i} = i^2[/tex], you had the idea that you could solve the equation [tex]e^{\pi \cdot x} = x^2[/tex] for x, and wondered if you would find other solutions besides i. (You can't solve an equation for the value of a constant, like solving 2(3) = 6 for 3, so the way you stated the question was confusing.) I'm not sure whether that equation can be solved at all; if so, it would involve some very ugly manipulations!
    I was trying to solve for x. I was trying to solve for i. i should be the square root of -1 however there should be no real answer to the square root of -1 so the square root of -1 is a different type of number which we call i. What I was trying to figure out is if there was way, through Euler's Identity, to find another solution to the square root of -1. Though it may be impossible like you said or that the results may just return to the square root of -1.

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    Quote Originally Posted by predarsh View Post
    I was trying to solve for x. I was trying to solve for i. i should be the square root of -1 however there should be no real answer to the square root of -1 so the square root of -1 is a different type of number which we call i. What I was trying to figure out is if there was way, through Euler's Identity, to find another solution to the square root of -1. Though it may be impossible like you said or that the results may just return to the square root of -1.
    There is only one "value" for i, which is the (imaginary) number called i. It is not a real number; you can't find its value by solving an equation until you have first defined i so that imaginary numbers can be talked about!

    I replaced i with x in your equation because you are looking for ANOTHER number besides i that satisfies the equation. As I said, you can't solve an equation for a constant; you can only solve for a variable, so we have to use a variable name. (You are perhaps thinking of taking i as the name of a variable; but that gets confusing. Changing the name of a variable does not change the algebra.)

    Bob Brown's link takes you to the solution of the equation you are asking about on Wolfram Alpha, a powerful computer program that can solve all sorts of equations (which knows what i is, so you couldn't reasonably tell it to treat i as a variable). It claims that the number -0.474541 is a real solution to the equation e^(pi x) = x^2, meaning that it could replace i in your equation e^(pi i) = i^2. And it is correct; I get

    e^(pi*-0.474541) = 0.225189159873...
    (-0.474541)^2 = 0.225189160681...

    But that does not mean it is a square root of -1. What has happened here is that there are many solutions to your equation, ONE of which is i; in producing your equation, you introduced extraneous solutions that are not solutions of x^2 = -1, which is the definition of i.

    I am curious as to why the program did not observe that another of the solutions is i; but that often happens. It is a machine, and doesn't know what your goal is, or even attempt to give all the solutions.

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