the problem, Create a class called Complex for performing arithmetic with complex numbers complex number have the form\nrealPart + imaginaryPart 8 i\nwhere i is the square root of -1.\nwrite a program to test your class. Use floating point variables to represent the private data of the class. Provide a constructor the enables an object of this class to be initialized when it is declared. \nProvide a no argument constructor with default values in case no initalizers are provided. Provide public methods that perform the following operations:\na)add two complex numbers\nb)Subtract two complex numbers\nc) print complex numbers in the form (a,b) where (a) is the real part and (b) is the imaginary part.\n\nAny help is GREATLY appreciated.
213639
Fairly straight forward. The code below is the skeletal C++ code (since I am not familiar with Java).\n\nclass Complex {\npublic:\nComplex () {a = 0; b = 0;}\nComplex (float x, float y) { a = x; b = y;}\nfloat real () {return a;}\nfloat imaginary () {return b;}\nvoid Add (Complex c, Complex d) { a = c.real() + d.real();\nb = c.imaginary() + d.imaginary();}\nvoid Subtract (Complex c, Complex d) {....}\nvoid Print() { cout << a << "+ i" << b; }\nprivate:\nfloat a;\nfloat b;\n};
169541
My computer gives me this 4 real results, there are two more Imaginary results.\n\n1) -1\n2) 1.8698\n3) -0.260\n4) -2.688\n\nImaginary roots are\n\n5) 1.039 + j1.40\n6) 1.039 - j1.40
342591
Start by putting the integer coefficient on the other side:\n\nx^2 - 12x = -7\n\nNow take the coefficient on the x-term (-12), divide it by 2 (-6) and square it (36).  Add this to both sides:\nx^2 - 12x + 36 = 29\n\nNow you can turn the left side into a perfect square:\n(x - 6)^2 = 29\n\nTake the square root of both sides:\nx - 6 = +/-sqrt(29)\n\nThen solve for x by adding 6 to both sides:\nx = 6 +/- sqrt(29)\n\nFor the second problem, it is the same method:\nx^2 - 5x = 11/4\nx^2 -5x + 25/4 = 11/4 + 25/4\n(x - 5/2)^2 = -14/4\nx - 5/2 = +/- sqrt(-7/2)\nx = 5/2 +/- sqrt(7/2) i\n\nThe only complications are if the coefficient of x^2 is not 1.  If it is something else, then just divide both sides by that at the second step to make it x^2.  Also, don't forget to look out for sqrt of negative numbers giving complex answers like #2.  I almost missed the sqrt(-7/2) = sqrt(7/2) i.  And you might want to simplify that so there is no sqrt in the denominator.
21064
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
342182
The sqrt (-9) = sqrt (9 * -1) = sqrt (9) * sqrt (-1) =3i.\nsqrt(-1) is defined to be i. Also sqrt(9) =3.\n\nYou should know that in general:\nsqrt(a*b) = sqrt (a) * sqrt(b).\nAlso sqrt (a/b) = sqrt(a)/sqrt(b)\n\nThe answer (3i) is a pure imaginary number becaus it has no real part. A complex number is one that has a real and an imaginary part such as 5+ 6i.\n\nHope this helped you.
519915
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.
523928
#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.
506147
It looks like you are trying to solve a literal equation, which is an equation with more than one variable, and you have to solve for a specified variable.  You can use the SAME rules to solve this kind of equation as you do with regular equations (those with numbers).  For example, if you had 8=2(b)(9), you would divide by 2 and then again by 9 (which is like dividing by 18).  \n\nIf your problem is a=(1/2)bc^2, then you would divide by 1/2 and then again by c^2.  Remember that to divide by a fraction, you multiply by the reciprocal.
412913
first of all, you need to know that there are basically two types of numbers. Real and imaginary. \nreal numbers are those which can be represented in one form or another, but they exist in real world. for example, if you say that there are 4 trees, that means that you can count the number of trees to be a distinct value, ie., 4.\nor consider this. the probability of an event happening is 0.58. this means that the event will be successful 58 times if it is performed 100 times. \nreal numbers consist of all the rational, irrational numbers. that means all these can be realised to be present.\n\nnow coming to imaginary numbers, these are not actually present. \nconsider a number like sqrt(-1). we all know that a square root of a negative number doesnt exist. so, we just say that sqrt(-1) doesnt exist. such numbers can be conveniently represented as Imaginary Numbers. these numbers donot mean anything in the real world, but help in our mathematical calculations.\nnow, hope this answers your question....
510625
it can be factored over the complex numbers. \nThe complex roots are:\n(1+i)/sqrt(2), (1-i)/sqrt(2), (-1-i)/sqrt(2), (-1 +i)/sqrt(2)\ni.e.\nx^4+1 = ( x-(1+i)/sqrt(2))( x- (1-i)/sqrt(2)) (x - (-1-i)/sqrt(2) )( x-  (-1 +i)/sqrt(2) )\n'
419929
If I got it, your problem is.........\n\n4+/- sq root -12\n_______________\n           2\n\n\n\nThen:   4+/- 2i sq root 3\n            _____________\n                  2\n\n\nSimplify to:    2+/-  i sq root 3
238934
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.
379534
I have to go teach my class in about 4 minutes, but when I return, if no one has done it, I can show you how.  It requires factoring out a specific value that makes the coefficients "a" and "b" turn into two values that satisfy x^2 + y^2 = 1.  In other words, I know it starts out by factoring out sqrt(a^2 + b^2) from \n\na sin(Bx)+b cos(Bx)\n\n\nto get \n\n\nsqrt(a^2 + b^2) [ a/sqrt(a^2 + b^2) * sin (Bx) + b/sqrt(a^2 + b^2) * cos(Bx)]\n\n\nEDIT: Okay here we go.  And the "orthogonal rotating vectors" proof below is usually over most people's heads, or at least, most people who would ask this question on Yahoo answers.\n\n\nFirst, you factor out SQRT(a^2+b^2) as I mentioned above.  For the sake of simplicity, I will let A = SQRT(a^2+b^2).  So factor A out of  \n\na sin(Bx)+b cos(Bx)\n\nto get\n\nA [a/A sin(Bx) + b/A cos(Bx)]\n\n\n\n\nSince (a/A)^2 + (b/A)^2 = 1 (you might need to actually verify this for your purposes), there exists some angle C such that\n\ncos (C) = a/A and sin (C) = b/A \n\n(also, you should point out that both a/A and b/A have absolute values less than one, so that C is real and not complex with a nonzero imaginary part).\n\nSo make the substitutions to get\n\nA [cos(C) sin(Bx) +  sin(C) cos(Bx)].\n\n\n\n\n\nAccording to the formula \n\nsin(y+z) = sin y cos z + cos y sin z, \n\nthe expression in the brackets must equal sin(Bx + C).  So now the whole expression becomes\n\nA sin (Bx + C).\n\n\n\n\n\nSo in conclusion, \n\na sin(Bx)+b cos(Bx) = Asin(Bx+C),\n\nwhere A = SQRT(a^2+b^2), and C = arccos(a/A).  \n\n\n\n\nThere is a better representation for C, however.  \n\nSince cos C = a/A and sin C = b/A, you have\n\ntan C = (sin C)/(cos C) = (b/A)/(a/A) = b/a.\n\nThus, tan C = b/a, and C = arctan (b/a).  This definition of C does not rely on the creation of a third variable (A).\n\n\n\n\n\nHope this helps!
425095
Answer:\nThe problem to solve is: \nRemember, with complex variables, keep like terms involving i together...\n \nAdding Multiplication Sign: y*x2-4x-5\nAdding Multiplication Sign: y*x*2-4x-5\nAdding Multiplication Sign: y*x*2-4*x-5\nPoly Parsing: y*x*2-4*x-5\nSplitting y*x*2-4*x-5 \ny*x*2-4*x - 5 \n[Left Side]Poly Parsing: y*x*2-4*x\nSplitting y*x*2-4*x \ny*x*2 - 4*x \n[Left Side]Poly Parsing: y*x*2\nMultiply Term y with Term x equals Term xy\nMultiply Term xy with Term 2 equals Term 2xy\nAddTerm: Term (2xy)\nPoly =  2xy\nRepace with Reduced Left Side \ny*x*2 - 4*x \n2xy - 4*x \n[Right Side]Poly Parsing: 4*x\nMultiply Term 4 with Term x equals Term 4x\nAddTerm: Term (4x)\nPoly =  4x\nRepace with Reduced Right Side \ny*x*2 - 4*x \n2xy - 4x \nBecause of the minus sign 4x becomes -4x\n\nPoly Adding 2xy with -4x...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\n... 2xy-4x\nPoly =  2xy-4x\nRepace with Reduced Left Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \n[Right Side]Poly Parsing: 5\nAddTerm: Term (5)\nPoly =  5\nRepace with Reduced Right Side \ny*x*2-4*x - 5 \n2xy-4x - 5 \nBecause of the minus sign 5 becomes -5\n\nPoly Adding 2xy-4x with -5...\nAddTerm: Term (2xy)\nAddTerm: Term (-4x)\nAddTerm: Term (-5)\n... 2xy-4x-5\nPoly =  2xy-4x-5\nThe problem to solve is: yx2-4x-5 \n2xy-4x-5
666374
Try solving problems and questions, watching quiz`s and reading (books or on line). Here`s one to get you started :\nwhat are the square roots of    -7 + 24i\npronounced square roots of minus seven plus twenty four eye.\n\nif you don`t know about complex numbers, research into this type of problem will increase your knowledge and hopefully open your mind up to new ideas and leave you wanting to find out more.
317938
the answer doesnt work out correctly because you have a negative under the square root of the quadratic equation.  but it can be solved using i.  then the answer would be 2.33i and -1.73i
85120
I think the point may be to use synthetic division.  Look that up in your text.