Absolute values of complex numbers?
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
141772
Also, remember that if the value inside the absolute value bars is negative, it converts to positive.  So to solve an equation like |x+3| = 5, you really have to solve TWO equations:  x+3=5 or x+3= -5.\n\nIf the absolute value portion of the equation is not by itself, you must make it so first.  For example:\n\n|2x + 1| + 3x = 5       First subtract 3x from both sides:\n\n|2x + 1| = 5 - 3x      Then split into 2 equations:\n\n2x+1=5-3x  or 2x+1 = - (5-3x) [which would be -5 + 3x]\n\nsolve each equation to get the 2 values of x
368083
Imaginary and Complex Numbers\n\nDo not let someone argue that all numbers are imaginary...just a side note.
453007
75: 8.660254038\n18: 4.242640687\n24: 4.898979486\n250: 15.8113883\n\nAll values are approximate
681247
Ordinary numbers can be represented as a line, like a ruler.   There's 0, 1 is to the right of it, 2 is further right, 1.5 is halfway between 1 and 2 and so on.   Complex numbers can be represented as points on a flat surface.   There's a real part like an ordinary number, and an imaginary part, which is a multiple of the square root of minus 1, written as i.   So 3-2i is the point you get to when you start off at the origin (like the origin of an x,y graph)You go 3 to the right and 2 down.   3+2i would be 3 to the right and 2 up.   Let's call a complex number z.   Turn z into (z times z) +1.   Call this z.   Do the same thing again, again and again, say 1000 times. z will either get bigger and bigger without limit or it will stay small.   The set of values of z that stay small is the Mandelbrot Set.   You can draw a picture of it.   If your piece of paper is white, and you make a black mark for every point on it corresponding to a complex number in the Mandelbrot Set, you get a black shape looking a bit like a peach with warts on it.   It's infinitely complicated; if you magnify the edge, you see more and more detail.
678825
The square root of -1\nis an imaginary 1\nit is depicted generally with a character 'i'\nif u want root of -2\nthen itd be i x root(2)\nor 1.414i
150881
There are several reasons. One of the most basic ones is that this representation provides a "polar" representation of complex numbers. Any complex number z=a+ib can be written as\n\nz= r exp(i theta)\nr= sqrt(a^2+b^2)\ntheta = atan(b/a)\n\nThis helps because, for example, a product of z1 and z2 would give\n\nz1 z2 = R exp(i phi)= r1*r2 exp (theta1 + theta2)\nR=r1*r2\nPhi = theta1 + theta2\n\nSo multiplying by z2 "stretches" z1 by amount r2 and "rotates" z1 by theta2. Such interpretations are convenient in many instances. Also, exponential functions are easier to deal with than regular a+ib type expressions in other cases.
403741
1.1 and -1.1\nbut you don't need to round it, that is the exact answer.
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
955805
That simplifies to 3[2x-1]>12\n\n[2x-1]>4\n\n2x-1>4    2x-1<-4\n\nx>5/2  or x<-3/2\n\nWhen you have absolute values, you have to look at both ends of the spectrum
419172
(0, 1)\n(1, 1.1)\n(2, 1.21)\n(3, 1.331)\n(4, 1.4641)
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
1169703
1. absolute value of any # is always positive.....so -7.5=7.5\n2. simplify, evaluate, solve....in math are pretty much the same things....in this case u only hav to bring out the 17 out of the absolute value. therefore ur ans wud b 17.\n3. devide -18/25, which is -0.72. since its in absolute value...the final ans wud b 0.72\nB. add 3+-4, which cud also b written as 3-4. the ans for that will b -4. since again its in absolute value, the final ans will b 1.
518082
additive inverse of any number 'a' simply means a number 'b' such that a+b=0, that is b = -a\n\nSo, if the complex number is 'p + iq', its additive inverse is: - p - iq\n\nIf you need to know abt multiplicative inverse, google on it, i am attaching a good link i know.
740420
Wait a minute everybody!\nthose vertical lines,( lyl ) could mean absolute value!!\nHelloooo?\nso the answerSSS could be 9 and -9\nsheesh!
1114980
A = pi * r^2\n\nIn this case r = c / 2pi = 8pi / 2pi = 4\n\nA = pi * (4)^2 = 16pi
654434
Yes, there is. The square root of negative 1 is the number i, which is an "Imaginary Number." \n\nDespite the name, imaginary numbers are real quantities that are used in equations to get real answers. However, they are a somewhat confusing concept, mathematically. \n\nOne way to visualize them is to consider that multiplication by an imaginary number rotates a function orthogonally on a number axis. The answers given by performing this procedure once may not seem to be applicable to real situatons, but if you multiply by an imaginary number again, in rotates the function back onto the initial axis, yielding real number results. \n\nI am including links to a number of good sites on imaginary numbers. Although it can be difficult to understand the implications of imaginary numbers beyond the fact that their squares are negative, if you spend some time on these sites, I'm sure you'll come to understand them.