what is the value of of i in the function e^(i*3.1416) = -1 ?

i got the value of

when i = 1 exp(3.1416) = 23.1407

when i = 3 exp(3.1416 * 3) = 1.2392e+004

when i = 45 exp(3.1416 * 45) = 2.4942e+061

and then how is e^( i *pi) = -1

what is the value of i in ths function?

10 Answers

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  • Remzy
    Lv 4
    1 decade ago
    Favorite Answer

    i is the square root of " -1 "

    in any differential equations book, you can find the proof of the following equation;

    exp (i*b) = cos(b)+i*sin(b) where " i " is the square root of " -1 "

    since in your function b=3,14

    cos(3,14)+i*sin(3,14) = -1+i*0 = -1

  • 1 decade ago

    The other answers are correct.

    In this case, i is not a variable, it is the usual letter used to represent the imaginary number which is the square root of -1, just like the Greek letter pi represents the irrational number that is the circumference of a circle divided by its diameter and the letter e is the base of the natural logarithm.

  • 1 decade ago

    Whoa there! You are treating i as if it were a variable to be solved for. It is not. It is a constant, equal to the number whose square is ,,, (-1). It is not part of the Real number set; a larger set of numbers, the Complex numbers, was devised to make use of it.

    Although it is not part of the number set known as Real numbers, it is very real in the sense that it is useful in real world engineering computations. It was invented several hundred years ago by mathematician C.F. Gauss so that every polynomial equation would have its proper number of solutions. When Gauss invented i, it was thought that i would have only theoretical interest, so it was called an "imaginary number", abbreviated i. Much later, i was found very useful in engineering math.

  • 1 decade ago

    You can immediately solve the

    problem if you know Euler's equation.

    e^iz=cosz+isinz

    you substitute the value of Pi for z

    and you get to result of -1

    so i=sqrt -1

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  • 1 decade ago

    'i' is in fact the square root of negative one. Many teachers of mathematics assert that this is not possible, but it comes about in real-world mathematical models.

    You'd need to take advanced courses in trigonometry and differential equations to really see how the relationship works.

  • 1 decade ago

    i=√-1 an imaginary number. That changes everything.

  • 1 decade ago

    i is the square root of -1

  • Anonymous
    1 decade ago

    i=sqrt(-1), so that i^2=-1 - imaginary unit

  • 1 decade ago

    i = square root of -1

    e^ix = cos x + isin x

    You misunderstand.

  • Anonymous
    1 decade ago

    e^(i*pi) = -1

    i*pi = ln(-1)

    i = ln(-1) / pi.

    There you have it! ;-)

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