Finite or infinite number of roots!
The equation x 3 = − 1 has three solutions, one of which is real and the other two are non-real complex numbers. Determine the number and type of solutions of x 2 1 = − 1
Note : When x is a complex number different from 0 , and r is a real number, x r can have more than one possible value. In this case, we assume that the complex number x is a solution of the equation x r = s , where s is a given real number, if at least one of the values of x r is equal to s .
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1 solution
I don't think this is correct. (try your solution with a computer) I think there is no solution, at least with the usual definition of complex exponentiation: The problem is that 2 π is not in the range ( − π , π ] . Complex exponentiation is defined by the principal value of the complex logarithm. In this case, it would be ( 2 − 1 ) π , which does not give the correct answer.
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Fabian, you know that e a + i b = e a ( cos b + i sin b ) for any pair of real numbers a and b . Then you should accept that − 1 = e ( π + 2 n π ) i and that the numbers e 2 ( π + 2 n π ) i are well defined complex numbers and they are are different from each other. If you accept all this, you should also accept that when you raise any of this numbers to 2 1 you get − 1 . Remember that the complex function a z is multi-valued, but you are talking just about a specific branch. If you enter the following expression in Wolfram Alpha: (e^(sqrt(2)(pi +6pi)i))^(1/sqrt(2)), you get -1. You can explore many other values by replacing the number 6 by any even number.
sir ,i have a doubt. suppose i am squaring both sides ,then i have, x^(sqrt 2)=1; i am taking this question as a combination of two function y=1 and y=x^(sqrt 2);now y=1 is a straight line parallel to X axis ,and y=x^(sqrt2) has value zero at x=0, also it is never negative, it is monotonically increasing function ,since dy/dx is always greater than zero. so graph of y=x^(sqrt 2) will always lie in 1st quadrant and it will surely intersect y=1 at one point ,so there should be only one real solution.
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Remember that whenever you square both sides of an equation, in general you get a new equation that is not equivalent to the original. That is, for example, the equation x = 1 has only one real solution and when you square both side you get x 2 = 1 with two solutions. Another example, x 2 = − 1 has no real solutions, but when you square both sides you get x 4 = 1 that has two real solutions. In general, when you square both sides of an equation, the set of all solutions of the new one contains the set of all solutions of the old one, but the two sets of solutions are not necessarily the same.
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thank you sir, thanks a lot. you cleared my doubt , will remember you for this problem. Regard- Manish
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@Manish Kumar Singh – Thank you, @manish kumar singh !
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@Arturo Presa – sir can i have your email address.
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@Manish Kumar Singh – My email is arturopresa@yahoo.com
Nice solution!
Wolframalpha input x^((sqrt2)/2)=exp ((pi +2 pi n)i)
= ( -e^(2 pi n i) )^( sqrt (2) )
We know that the number − 1 can be represented in the form e ( π + 2 n π ) i , where n is an arbitrary integer number. Then the given equation can be written in the form x 2 1 = e ( π + 2 n π ) i . Raising both sides to 2 , we obtain that x = e 2 ( π + 2 n π ) i ( ∗ ) . Since 2 is irrational then the numbers 2 ( π + 2 n π ) i , where n is an integer, are always different from numbers of the form m π i where m is any integer, and the differences of any pair of those numbers are always different from a number of the form 2 m π i where m is an integer. Therefore, the number of possible values of x given by the equation ( ∗ ) is infinite and none of them is real.