Ok so we know e^(pi*i)=-1      what about this?
820932
Logarithms, when applied to imaginary numbers can have many results. This is because exponentiation, when applied to imaginaries, is defined as a trig function:\n\ne^(ix) = cos x + i sin x\n\nSo, the logarithm is an inverse trig function. And we know inverse trig functions can have many results. ln e^(2 pi i) can be 2 pi i, but it could also be 0, or -2 pi i, or basically any ix where (cos x + i sin x = 1). In this case 0 fits with the real logarithm on the right side, since ln(1) = 0.
752580
It just takes foiling-\n\n(5*2)+(5*5i)+(-3i*2) +(-3i*5i)\n(10)+(25i)+(-6i)+(-15*i^2)\ni^2=-1\nso...\n25+19i\n\nTo those who have said that you can't simplify it past 10 + 19i - 15i^2, that isn't correct considering that since i is the square root of -1, i^2 is -1. That's why you can simplify -15i^2 to 15, thus making it 10+19i+15, or 25+19i
837906
x is (2n+1)th root of 1\n\nso x = e^(kpi/(2n+1)) k is from 1... 2n\n\n1+x+x^2+....+x^2n = 0\n\nx+x^2+.... + x^2n = -1\nthe real part of this is \n\ncos(2pi/(n+1))+cos(4pi/(n+1)+ ... = -1
706332
The antiderivative is:\n\n(1/2)erfi(sqrt(pi)*x)\n\nI used Wolfram
24161
The square root of a negative number is an imaginary one.  The square root of -1 is known as i.  Negative square roots work out like positive ones, only with the i, so SqRoot(-4) = 2i and SqRoot(-64) = 8i.
276671
(e^(cose^e+sin^e)/2)^pi\nhahaha ..just kidding
504832
f(pi/6) = cos(3pi/6)\n= cos(pi/2)\n= 0 a.\n\nRoots of x(x^2-1)(x^2+1) are 0, 1, -1, i, and -i, so there are 3 real roots.\n\ny = e^x - 1\ne^x = y + 1\nx = ln(y+1), y > -1\n\nAlthough if you have multiple questions, then you should post separately.
808989
Let t=sinx\n=> dt = cosx dx\n\nx=pi/2 => t=1\nx=0 => t=0\n\nI=integrate(e^t dt) from 0 to 1\n\n=e^t from 0 to 1\n\n=> I= e^1 - e^0 = e-1\n\nAnswer is I=e-1
179378
All you do is take the natural logarithm (ln) of both sides.\nBecause the ln of e to any power is just that power it becomes this:\nln (e^-x) = ln(6) \nThe ln sort of "cancels out" the e, leaving you with the power so...\n-x = ln(6)\nMultiple both sides by negative 1 and you're done, unless you need to approximate ln(6).
607947
i^-1= 1/i^1  you can get rid of the negative exponent by changing it to the reciprocal.
702351
I think you are looking too much into it, but here we go. \n\nIf you have the equation x^2 and say we pick points x= -1, and 1. \n\nWhen you plug the x's into the equation you will get a positive answer. This is why whenever the exponent is even, it will always bounce off the x- axis, because there is no way for it to go negative.\n\nfor the second question, say we take x^2 and plug in the same points above.\n\n-1 ^ 3 gives you -1. This would mean that ever x = -# will have a negative y value which means it will cross the x- axis\n\nTry it with more equations and you will understand
511125
Its either d or e. It should be a deshielding effect by the pi electrons. There is some gobbledy-gook showing up in my browser so I can't tell d or e is the pi symbol.
6015
intgral(from - infinity to infinity) of e^(-x^2)= sqrt(pi)\n\nThis is the area under the bell curve. It comes up very often in statistics.
200519
+/- 1\n\nThe answer is not 1, because -1 * -1 = 1 also.
559973
4i is the answer you seek,\nthe i is the square of -1 says any geek.\nThis is a abstract answer to solve your query,\nbut all negative roots include the 'i,' my deary.
821031
I=I1+I2; I1=int{for x=pi/6 until pi/2} of x*dx = x^2/2{for x=pi/6 until pi/2}= pi^2/9;\nI2=int{for x=pi/6 until pi/2}of  25*dx/(sinx)^2; \nLet t=tg(x), then 1/(sinx)^2=1/t^2+1 and dx=(1+t^2)*dt;\nI2=int{for t=1/sqrt(3) until +inf}of 25*t^(-2)*dt = -25/t = 25*sqrt(3)
1156569
The problem is not that you can't take the log of a complex number. The problem is that log is multiple valued in the complex plane. In other words, the exponential function is not one-to-one. You can take the 'principle branch' of the logarithm, but then you can lose certain of the rules of logarithms.\n\nThe upshot is that when you take the log of -1, you either get an infinite set of values (.., -pi*i,  pi*i, ...) and another infinite set of possibilities for the log of 1 OR you lose the property that log(a^b)=b*log(a) that you used in your first paradox. This is just a fact of life when working with complex numbers. In the second, BOTH -pi*i and pi*i are logarithms of -1, so the argument fails. It's somewhat like saying that the squares of -1 and 1 are the same, so -1=1.