Anonymous asked in Science & MathematicsMathematics · 9 years ago

hey need help with math project?

hello everyone i have a math project to increase my grade,so i want to ask where do we use complex numbers in real life i need clear anwears cuz my math isnt good to understand complicated details.i also want to know where and how it is exactly used.also can you prefer me a music for my slide must be more science music.

thanks for all your helps



well ok i need where we use it everyday,in real life,for example in Technology and how we use it which branch of it

Update 2:

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5 Answers

  • Ed H
    Lv 4
    9 years ago

    Complex numbers enter into studies of physical phenonomena in ways

    that most people can't imagine. There is, for example, a differential

    equation, with coefficients like the a, b, and c in the quadratic

    formula, that models how electrical circuits or forced spring/damper

    systems behave. The movement of the shock absorber of a car as it goes

    over a bump is an example of the latter. The behavior of the

    differential equations depends upon whether the roots of a certain

    quadratic are complex or real. If they are complex, then certain

    behaviors can be expected. These are often just the solutions that one


    In modeling the flow of a fluid around various obstacles, like around

    a pipe, complex analysis is very valuable for transforming the problem

    into a much simpler problem.

    When everything from large structures of riveted beams to economic

    systems are analyzed for resilience, some very large matrices are used

    in the modeling. The matrices have what are called eigenvalues and

    eigenvectors. The character of the eigenvalues, whether real or

    complex, is important in the analysis of such systems.

    In everyday use, industrial and university computers spend some

    fraction of their time solving polynomial equations. The roots of such

    equations are of interest, whether they are real or complex.

    And complex numbers are useful in studying number theory, which is the

    study of the positive integers. The techniques in complex analysis

    are just one more tool that researchers have.

    Source(s): Math Forum
  • ??????
    Lv 7
    9 years ago

    Well complex numbers are often connected with polynomial equations.

    We need to use complex numbers to solve cubic equations analytically example given.

    This means that we find exact solutions in terms of radicals. When a cubic equation has three different real values as solutions the substitution of Vieta needs complex numbers in it' s process to solve.

    So now we still need to find how we need cubic equations in real life. The cube of a value is often connected with volume, as volume is 3-dimensional. So we might say, search for a sphere that has as sum of the volume and the area (surface) 1.

    => (4/3) pi r³ + 4pi r² = 1

    => r³ + 3 r² - (3/(4pi)) = 0

    => r³ + 3 r² - 0.2387324146 = 0

    Solving this cubic equation with Vieta's substitution involves complex numbers, even when the solutions are all real !!! I will show you the process

    The Rational Roots Theorem gives us a clue on searching for rational roots :

    (searching for roots in the form +- p/q with p=divisor of 2387324146 and q=divisor of 10000000000)

    No rational roots.

    There is a method to solve a cubic equation in general by hand (and calculator) on paper.

    Dividing by the first coefficient yields :

    x^3 + 3 x^2 - (1193662073/5000000000)

    Substituting x=y+p in x^3+ax^2+bx+c yields :

    y^3 + (3p+a) y^2 + (3p^2+2ap+b) y + p^3+ap^2+bp+c

    if we take 3p+a=0 or p=-a/3, the first coefficient becomes zero, and we get :

    y^3 - 3 y + (8806337927/5000000000)

    (with p = -1)

    Substituting y=qz in y^3 + b y + c = 0, yields :

    z^3 + b z / q^2 + c / q^3 = 0

    if we take q = sqrt(|b|/3), the coefficient of z becomes 3 or -3, and we get :

    (here q = 1.00000000)

    z^3 - 3 z + 1.76126759

    Substituting z = t + 1/t, yields :

    t^3 + 1/t^3 + 1.76126759

    Substituting u = t^3, yields the quadratic equation :

    u^2 + 1.76126759 u + 1 = 0

    We have bad luck. The roots of the quadratic equation are complex.

    A root of this quadratic equation is u=-0.88063379 + 0.47379756 i

    Substituting the variables back, yields :

    t = cuberoot(u) = 1.00000000(cos(-0.88266485)+i sin(-0.88266485)) = 0.63509498 + -0.77243405 i.

    z = 1.27018996 + i 0.00000000.

    y = 1.27018996 + i 0.00000000.

    x = 0.27018996 + i 0.00000000.

    The other roots can be found by dividing and solving the remaining quadratic equation.

    The other roots are real : -2.97299001 and -0.29719995.

    Of course cubic equations are involved in lot's of other optimization problems and mechanical calculations !

    Hope this helps !

  • 9 years ago

    Complex Numbers are used to analyze properties of electronic circuits.

    Here is a link that explains a little bit about the application of complex numbers.

  • 4 years ago

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  • 9 years ago

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