2007-09-25 04:20:48 · 6 answers · asked by fishburn7 2 in Science & Mathematics ➔ Mathematics
e^x > 0 in real domain. Therefore, e^x = -1 is impossible in real domain.
In complex domain, e^(i*pi+2n*pi) = -1, where i is an imaginary number and n can be any integer.
2007-09-25 04:27:19 · answer #1 · answered by sahsjing 7 · 0⤊ 1⤋
No real number x satisfies this equation.
In the complex numbers, every number of the form k * Ï * i, where k is odd, satisfies this equation.
2007-09-25 11:28:33 · answer #2 · answered by Anonymous · 1⤊ 0⤋
e^x is always positive:therefore, there is non value for x that makes e^x= -1/
2007-09-25 11:33:23 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
x = i*(pi) worked out as under.
e^x = -1
=> e^x = cos(pi) + i sin(pi) = e^ (i*pi)
=> x = i*(pi), i = sqrt (- 1)
I saw Pascal's answer which is more perfect and hence has given him thumbs up. It would have been better if he would have shown steps.
cos(pi) + isin(pi)
= cos(pi + 2k*(pi) ) + si(pi + 2k*(pi)) = e^ (pi)*(2k + 1), k belongs to R.
2007-09-25 11:28:01 · answer #4 · answered by Madhukar 7 · 2⤊ 0⤋
iÏ(1+2k) for any integer k.
2007-09-25 11:27:43 · answer #5 · answered by Pascal 7 · 1⤊ 0⤋
pi*i
pi=3.141592653...
i = sqrt(-1)
2007-09-25 11:26:59 · answer #6 · answered by Love Zelda0 2 · 1⤊ 0⤋