What is i^i (where i is the square root of -1)?
83972
R*exp(i*theta) = R [ cos(theta) +i*sin(theta) ]\nthis is the polar representation (raduis R ,angle theta) transformed to the cartisienne form (X=R cos(theta), Y=R sin(theta))  where i is imaginary to matk that y is perpendic to x\n\nR=1 theta=pi/2\nexp(i*pi/2) = cos(pi/2) + i sin(pi/2)\nexp(i*pi/2) = 0 +i *1 =i\n\nso i=exp(i*pi/2)=e^(i*pi/2)    where e=2.7818281828\ni^i=e^(i*pi/2)^i\ni^i=e^(i^2*pi/2)\ni^i=exp(-pi/2)\n=0.207879576
681679
The solution involves complex numbers ('i' notation)\n\nThe factorized equation is:\n\n(b + 3 - i)(b + 3 + i)\n\nwhere 'i' is the imaginary complex number = square root of -1\n\n(TOOK ME WHOLE OF 15 MINS!!! MY MATH IS SLIPPING!!!)\n\neinsteinbrainiac - you came close to the solution
456629
The absolute value of the complex number x + iy is defined as sqrt (x^2 + y^2).\nSo,\n1) |2 + i| = sqrt(2^2 + 1^2) = sqrt 5\n2) |3i - 2| = sqrt(3^2 + (-2)^2) = sqrt 13
705926
let cube root of 1 be x. \n=> x^3 = 1\n=> x^3 - 1 = 0\n=> (x-1) (x^2 + x + 1) = 0\n=> x=1 or x^2+ x+1=0\n=> x=1 or x= [1+ i(3)^1/2 ] /  2  or  [1- i(3)^1/2 ] /  2 \n  where i is iota, (-1)^1/2\nHence cube root of one has three roots, one real and 2 imaginary. these imaginary roots are called w (omega) and omega square. \n1+w +w^2 =0\n1* w * w^2 = 1
786717
its 1\n\nsince i = (-1)^1/2\n\nso i^2 = (-1)^(2*(1/2))\n\nwhich is 1
211940
This is your fifth question in math...Do you find the subject difficult?...Ok...If your question is correct...\n\n3x^2 + 13 - 10 = 0\n\n3x^2 + 3 = 0\n\n3x^2 = -3\n\nx^2 = -3/3\n\nx^2 = -1\n\nx^2 = i^2 ===>i is a symbol in complex numbers\n\nwhere i^2 = -1. \n\nx^2 - i^2 = 0\n\n(x + i)(x - i) = 0\n\nTherefore:\n\nx + i = 0 ......... x - i = 0\n\nx = -i ............. x = i ANSWERS or\nx = - sq. root of -1..... x = sq. root of -1 ANSWERS\n\nThe solution is imaginary and NOT real.
1122465
Just plug in 1 for x.\n√(1-1) = √0 = 0
966844
f(0) = -i\nf(1) = 1² - i = 1-i\nf(2) = 2² - i = 4-i\n\nI take it you are dealing with complex numbers?  i is the square root of -1.\n\nNRT
517182
I don't think it is possible to actually comprehend i, since it is the square root of -1.  We don't know of any numbers whose square equals a negative number, which is why i is imaginary.  When you square a negative number, the negative signs cancel out, giving a positive value.  In order to get a negative square, you would have to have some sort of crazy negative/positive amalgam or some type of half-positive-half-negative number.\n\n\nimaginary numbers are ones that aren't possible in our normal, continuous mathematics.  They are virtually always represented with i, since i is the basis for imaginary numbers.\n\nI would say that yes, you can have negative somethings, even though you have to imagine them (which is why "imaginary numbers" might be a misnomer).  For example, if you have six apples but you are hungry, you might think of your hunger as -2 apples.
435410
2i or -2i, where i = sqrt(-1)\n\nIf you haven't learned imaginary numbers, then your answer would be that the sqrt(-4) is not a real number.
866033
1 followed by 10^50 zeros\n\nSince a googolplex (notice it's googol, not google like the search engine) is a 1 followed by 10^100 zeros, the square root of that is just half the exponents; just like sq.rt. of x^2n = x^n.\n\n~ ♥ ~
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.
206636
for (i=1,t=1,cosx=1;i<=n;++i)\n{\n  t=(-1)*(t*x*x)/((2*i)*(2*i-1);\n  cosx += t;\n}
770326
If you are allowed to head into the realm of complex/imaginary numbers, you could factor out a sqrt(-1) and replace it with a lowercase i.\n\nsqrt(-24) = sqrt(4) x sqrt(6) x sqrt(-1)\nsqrt(-24) = ± 2 sqrt(6) i
948937
(-3,-3);(-2,0);(-1,1);\n(0,0);(1,-3)
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.
230430
The square root of -1 has no relation to zero.  It requires an imaginary number.  Division by zero is, unfortunately for our number system, impossible.  You cannot cut something into zero pieces, and in fact the very statement makes no sense.