how do u find the number of imaginary zeroes using Descartes' Rules of Signs?
108340
Descartes' Rules of Signs is used to find the number of positive and negative REAL zeroes only.  Thus, however many roots are left after the real roots must be imaginary.  By the way, imaginary roots come in conjugate pairs, therefore you will always have an even number of imaginary roots.
704203
we use imaginary number in solving equations where there are 2 components of a thing like in solving electrical circuits
726208
I think the easiest way to know whether or not you add is by picturing a number line that is negative on the left and positive on the right with zero in the middle.\nif you subtract, then you move left\nif you add, then you move right.\n\n\n\nMemorize this!\n+ + makes +\n+ - makes -\n- + makes -\n- - makes +\n\nexample:\n1+(+2)=1+2\n1+(-2)=1-2\n1-(+2)=1-2\n1-(-2)=1+2\n\nanother example:\n(-1) + (+2) = (-1) + 2 = 2 - 1\n(-1) + (-2) = (-1) - 2 = -(1+2)\n(-1) - (+2) = (-1) - 2 = -(1+2)\n(-1) - (-2) = (-1) + 2 = 2 - 1\n\ntry these, and we'll tell you if you got it\n4 - (-5) =____\n6 - 7 = _____\n-9 - 8 = ______\n-3 - (-5) = _______
131038
Something is screwy with your question.  The second part is just a transformation of Euler's identity, e^ix = cos(x) + i sin(x)\n\nSo, your second expression simplifies to [e^(i 3pi/2)]/8\n\nIf I assume, like the previous poster, that your first expression should be (1+1/2i), then, just doing the algebra the left side of your equation gives, (1+1/2i)^6 = -117/64 - 11 i/16.\n\nFor the right side of your equation, for real, 1/8 cos 3pi/2 = 0\nFor imaginary, 1/8 i sin 3pi/2 = - i/8\n\nNow comparing left and right real and imaginary parts:\n0 != -117/64 and\n-i/8 != - 11 i/16\n\nSo, they are not equal.
83855
y = a*x^2 + b*x + c\n\nThat is, y = constant times x squared plus constant times x plus a third constant. The solution is Eqn 5 on the link below.\n\nIf the stuff under the radical in the solution is negative then you will have an imaginary number on your hands.
915238
if the number is positive you get real roots\nso root 36=+/-6\nif the number is negative you get imaginary roots\nso root -144=+/-12i
679535
The best way to approach this is to use logarithms & Euler's formula. \n\nWe know that the Euler's formula states:\n\nexp(ix)=cos(x)+i*sin(x)\n\nWe have to express 'i' in terms of the angle 'x' - to do this, we use argand diagram & find that 'i' is at an angle of 90 degrees (pi/2) to the Real axis. Hence, x=pi/2.\n\ni = exp(i*pi/2) = 0+(i*1) = i (which proves the last statement)\n\nNow, going back to the actual problem,\n\nA=i^i\n\nTake logarithm on both sides\n\nln(A)=i*ln(i)\n\nBut we have proved\n\nln(A)=i*ln(exp(i*pi/2))\n\nSince ln & exp are inverse of each other, they cancel out to give:\n\nln(A)=i*i*pi/2\nln(A)=-pi/2\nA=exp(-pi/2)\n\nThus, i^i = exp(-pi/2)
938958
No, it would have to be the square of something like ax+b, whose square is a^2x+2abx+b^2.  You would have to have a=1 or -1 and b=i or -i, but then 2ab would be imaginary, and not -2.
944309
#1.\n\nx^3 - x^2 + 16x - 16 = 0\nx^2(x - 1) + 16(x - 1) = 0\n(x - 1)(x^2 + 16) = 0\nx = 1, x = 4i, x = -4i\n\n4i is an imaginary number where i is equal to the square root of -1.  If you square i you get -1.\n\n#2.\nSet the denominator equal to 0.\n\n4x - 2 = 0\n4x = 2\nx = 1/2\n\n#3.\n8x - 12\n---------\n4x - 2\n\nThe limit as x approaches infinity equals 2.  The -12 and -2 are insignificant as X approaches infinity so all you have left is 8x and 4x.  If you divide 8x by 4x you get 2.
807333
Imaginary numbers, actually complex numbers, are enormously useful in many areas of engineering.  One example is in trigonometry where difficult trigonometric identities become almost trivial to derive.  Another is in analyzing circuits.  Using imaginary numbers allows Ohm's law to be applied to passive (resistors, capacitors and inductors) AC circuits.  Euler's identity allows cubic equations to be solved analytically.  Applying the concept of graphical representation of 2-D vectors provides very powerful tools for the solution of 2-D kinematics problems and difficult geometry problems.\n\nI am a research engineer and I use complex numbers all the time in my work.  It is by far one of the most useful tools that I have to solve difficult problems.
1052440
The trivial zeros are at (-2K+0i), where K is a positive integer. The nontrivial zeros are all the others. It has been shown that the nontrivial zeroes must have a real part between 0 and 1.
89652
The formula involves complex maths calculations. Ask some maths professor.
815664
Imaginary part of\n(e^(i*x)+ e^(2*i*x) + ...)\n=Imag(e^(i*x)/(1- e^(i*x))\n=Imag(e^(i*x)/\n  (1-cos(x)-i*sin(x)))\n=Imag(e^(i*x)\n   (1-cos(x)+i*sin(x))/\n   ((1-cos(x))^2+(sin(x)^2))\n=((cos(x))\n(sin(x))+(sin(x))\n(1-cos(x)))/(2-2*cos(x)\n=sin(x)/(2(1-cos(x)))
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.
1115504
the cube root can be found using a calculator\nmake sure you have a scientific one though\ni had do use it alot in geometry\n\nits alot like a square root just three instead of 2
316406
Depends on how fast you're counting...\n\nhttp://hub1.worlded.org/docs/maththing/ny1p5.htm\nhttp://mathforum.org/library/drmath/view/58739.html
683943
Any complex number can be expressed as r*e^(i*theta).\n\n2^i = r*e^(i*theta)\n\nlogs on both sides\n\nln(2^i) = ln (r) + ln (e^(i*theta))\ni*ln(2) = ln(r) + i*theta\n\nequate real and imag parts:\n\nln(r) = 0\nso\nr=1\n\ntheta = ln(2)\n\nso the answer is 2^i = e^(i*ln(2))\n\nThere's an identity to do that more easily, but I can't remember it - it's something like this:\n\na^b = fn(e^b)