about complex numbers,how can i solve this?
953689
A clue:\n\nsin h x = (e^x + e^-x) / 2\n\nCalculate e^(2-i) = e^2 e^-i, etc\n\nsin h x = (e^2 e^-i +e^-2 e^î)^2\n\nAna
454578
mrniceguy's answer is correct, but he has one little eror in the work.  i^7 = -i
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.
638676
1.)\nstep one: if they are zeroes, use opp. \nexample....\n (x-2) (x+9) (x-7)          if you solve for x it should be opposite of given zeroes.\n\nstep two: FOIL the first two...\nyou'll get   x^ +9x -2x -18 which simplifies to\n\n(x^2 + 7x -18) (x-7) \n\nstep three: FOIL these two together...\nyou'll get     x^3 +7x^2 - 18x -7x^2 -49x + 126\n\nwhich simplifies to  \n   x^3 - 67x +126\n\n\n2.) I don't know about the second one... I could solve it if I spent some time, but start with...\n (x - 4i) (x+5) (x + 2i)\nthan just do what I did in the problem above, factor the first two, than factor your answer with the last one... and remember that i^2 is equal to 1...
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.
968624
1.  \n\n7x + 5i = 7 - 6yi\n\nIsolate x:  \n\n7x = 7 - 6yi - 5i\n\nSolve the right hand side in the proper form of complex numbers (a + bi)\n\n7x = 7 + i(-6y - 5)\nx = 1 + ( (-6y - 5)/7 ) i\n\nYour goal is to get some value for a (the number without the i) and some value for b (the number next to the i).  That's why I factored out the i so that we get the complex number proper form.\n\n2.\n\nThe quadratic formula for ax^2 + bx + c = 0 goes as follows:\n\nx = (-b +/- sqrt(b^2 - 4ac))/2a\n\nIn the above question, a = 2, b = 7, and c = 6.  All you need to do is substitute.  Note that "+/-" means "plus or minus", meaning you'll get two solutions for x.
1108271
=1/(z^2)-3z-4 + 3z/(z^2)-6z-7\n=1/(z^2)*(z^2)-3z*(z^2)-4*(z^2)+3z/(z^2)*(z^2)-6z*(z^2)-7*(z^2)\n=1-3(z^3)-4(z^2)+3z-6(z^3)-7(z^2)\n=1+3z-4(z^2)-7(z^2)-3(z^3)-6(z^3)\n=1+3z-11z^2-9z^3\n\nbascially multiply everything by (z^2) and then add the common coefficients
959491
sin(X)+sin(2x)+sin(3x)+....+sin(nx)\n= img[exp(ix)+exp(i2x)+exp(i3x)+...+exp(inx)]\n=img[(exp(i(n+1)x) -1) / (exp(ix) - 1)]\n=img[(exp(i(n+1)(x/2)) -1) /exp(i(n+1)(x/2)) * (exp(i(n+1)(x/2)) -exp(-i(n+1)(x/2))) / (exp(i(x/2)) - exp(-i(x/2)))\n=[sin((1/2)(n+1)x)/sin((1/2)x)]* img(exp(inx/2))\n=[sin((1/2)nx)/sin((1/2)x)]*[sin((1/2)(n+1)x)/sin((1/2)x)]/[sin((1/2)x]\n\n[exp is exponential, img is imaginary part, 'i' is usual imaginary i in complex numbers] I also suggest you write down this so that you will be more clear with the equations. Sorry but without equation editorits bit difficult to type math.\n\nIf you want an infinite sum, please take the limit as x-->infinity of the last term obtained.\n\nI hope this answeres your question. All the sest.
89652
The formula involves complex maths calculations. Ask some maths professor.
532355
Please see the method explained by me as an answer tothis question..\nhttp://in.answers.yahoo.com/question/index;_ylt=AiNJtRa.xTUMUTdP29hR7sSQHQx.?qid=1006051102360\n\nThis will help u solve cubic as well as other equations.. :-)
543826
(x+2)*(x-5i) = x^2+(2-5i)x-10i\n\nsomehow, i dont like these complex #'s\n\nyeah, the answer below mine is probably correct.\n\n(x+2)*(x-5i)*(x+5i) = ...
810486
Hi Arman! With no L’Hopital again? \n((x^1/4)-(x^1/3))/((x^1/2)-1) for x->1, mind! x>0\nLet z=x^1/12, then (z^3-z^4)/(z^6-1) =\n= z^3*(1-z) / {(z^3-1)*(z^3+1)} = \n= z^3*(1-z) / {(z-1)*(z^2+z+1)*(z^3+1)} = \n= -z^3/{(z^2+z+1)*(z^3+1)} = \n= /for z->1/ = -1/(3*2)=-1/6
525611
Imaginary numbers are just another word(s) for complex numbers. They are the same thing. Complex-Imaginary numbers are numbers that have a part of them that is multiplied by i (the square root of -1).\nEg 1+1i is a imaginary number\n(1+1i) squared = 0+2i = 2i\nso one of the square roots of 2i is 1+1i (a complex number)\nI use complex numbers to design fractal generation programs.
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
567357
I can tell by the way you asked the question that you probably already know the answer. Right?  \n\nI like how you phrased the question.  "Got a 'complex' question...?"\n\nThe set of complex numbers are made up of Real and Imaginary numbers. Real numbers are numbers that we use in everyday life. Where as complex numbers could include imaginary numbers, such as the square root of a negative number.\n\nDid you know that a mathematician by the name of Leonard Euler (pronounced "oiler") was the one who came up with the imaginary number "i"? which represents the square root of a negative number? He also came up with the number "e" for 2.718...
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
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.