Math problems help me plz?
Think of (x, y) points in the Cartesian Coordinate System as representing the complex number x + iy. (All real numbers are on the x axis. All complex numbers with the real part as equalling zero are on the y axis.) 2 + i is represented by the point (x, y) = (2, 1). If you connect (0, 0) and (2, 1) with a straight line, then the angle between this straight line and the positive x-axis is 30 degrees. Real part is positive -- point is to the right of the y-axis. Complex part is positive -- point is above the x-axis. That should get you started.
Math problem help! (Complex numbers)?
a) You are correct. It's easy to check that z = 1 is a root of z^5 - 1 = 0, because 1^5 - 1 = 0. So, (z - 1) is a factor of z^5 - 1. ==> z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1) by long/synthetic division. b) I'd use DeMoivre's Theorem. z^5 = 1 = cos 0 + i sin 0. ==> z^5 = cos(2πk) + i sin(2πk). ==> z = cos(2πk/5) + i sin(2πk/5) for k = 0, 1, 2, 3, 4. So, the five roots are k = 0: cos 0 + i sin 0 = 1 k = 1: cos(2π/5) + i sin(2π/5) k = 2: cos(4π/5) + i sin(4π/5) k = 3: cos(6π/5) + i sin(6π/5) k = 4: cos(8π/5) + i sin(8π/5) c) To do this efficiently, note that we can rewrite the last two roots as cos(-4π/5) + i sin(-4π/5) = cos(4π/5) - i sin(4π/5) cos(-2π/5) + i sin(-2π/5) = cos(4π/5) - i sin(2π/5) So, these are complex conjugates of the roots for k = 1, 2. Hence, z^5 - 1 = (z - 1)(z - (cos(2π/5) + i sin(2π/5))(z - (cos(4π/5) + i sin(4π/5)) * (z - (cos(4π/5) - i sin(4π/5)) (z - (cos(2π/5) - i sin(2π/5)) = (z - 1) * (z - (cos(2π/5) + i sin(2π/5))(z - (cos(2π/5) - i sin(2π/5)) * (z - (cos(4π/5) + i sin(4π/5)) (z - (cos(4π/5) - i sin(4π/5)) = (z - 1) (z^2 - 2z cos(2π/5) + 1) (z^2 - 2z cos(4π/5) + 1). I hope this helps!
Problem with Polynomials and Real numbers. Help!?
1) What is the actual meaning of polynomial? In mathematics, a polynomial is an expression consisting of variables (or indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. An example of a polynomial of a single indeterminate (or variable), x, is x2 − 4x + 7, which is a quadratic polynomial. 2) What is the meaning of p(x)? It is just a function of x, the value of p is determined by x 3) What is meant by the factors of a polynomial? After factorizing a polynomial, why do we equate it with zeroes? Solve quadratics by factoring, completing the square or Quadratic Formula -b/2a ±√(b² -4ac)/2a. We factor because when two items are multiply, like AB=0, then either A=0, B=0 or both. 4) What is meant by the roots or zeroes of a polynomial? Where the equation has the value y=0, crosses the x axis. 5) Why do we equate a polynomial equation with 0 in order to find its roots? That's where y=0, where it crosses the x axis. In the attached image of f(x) or y=x² -4, the roots are 2 & -2 because 0 = x² -4 factors are (x+2)(x-2)=0, then (x+2)=0 so x=-2 (x-2)=0 so x=2 . 6) The more the degree the greater number of roots will be (of a polynomial). Why? WHAT?? Real vs Imaginary: Imaginary is a number with √-1, otherwise they are real. Imaginary are JUST AS IMPORTANT as Real!! In fact, sometimes they are more useful. They need a better name!
What is the similarity between complex numbers [[math]i=\sqrt{-1}[/math]] and the whole concept of God being real or imaginary?
Some complex numbers are real and some of them are imaginary. Some are neither. The only complex number that is both real and imaginary is zero.God, on the other hand, is real to some people and imaginary to others. To some others, God is neither. Perhaps there are a few more people who perceive God as being both real and imaginary, but I haven’t met any such persons yet.Is that a similarity between complex numbers and God, to you? Yes? Then there you go: you have your similarity. No? Then sorry to disappoint you: there is no other similarity between complex numbers and God.
What is the minimum number of real zeros possible for a cubic polynomial with real coefficients?
By the Fundamental Theorem of Algebra, a cubic polynomial has 3 roots in the complex numbers. But since the polynomial has real coefficients, all complex roots must come in conjugate pairs. So the polynomial has either (i) 3 real roots or (ii) 2 complex roots and 1 real root. Thus the minimum number of real roots is 1. Note also that in the example above me, f(x) = x^3, the polynomial has 3 real roots; they just happen to be equal to each other.
Can you figure out two complex numbers (where neither a nor b are zero) that when multiplied together become a?
Let's say one complex number = a + bi and another = c + di. When multiplied together they give (a + bi)(c + di) = ac + adi + bci + bdi². Now recognising that i² = -1, this can be written as ac - bd + (ad + bc)i. If this is real, then ad + bc = 0, or bc = -ad. Let's take, for an example, a = 3, b = 6, c = 1 and d = -2. Then the complex numbers they represent are 3 + 6i and 1 - 2i. They will multiply together to give ac - bd = 3x1 - (-6x2) = 3 + 12 = 15.
What problems are hard (but not impossible) to solve without complex numbers?
In circuit analysis, if you have a complicated system involving a capacitor, a resistor, and a battery, and you want to figure out how the current flow, you can do it by setting up a differential equation. This will involve no complex numbers, but is an exceptional pain. If I recall correctly, when I took circuit analysis, out of a class of at least 10 or 15 students, no one managed to set up and solve such an equation correctly. (To be fair, none of us had taken a class on differential equations at that time, but we had a rough idea of what we were supposed to do.)Alternatively, you can pretend that you have real and imaginary current, work out how a capacitor and resistor work separately, and then show that if you have put them together, the system works in an essentially additive fashion. I don't remember the details any more, but I do remember how vastly easier it was to solve the problem by accepting the existence of complex numbers.
What is a complex zero in terms of Algebra?
As a follow up to User-13166949966061750191 answer, complex roots were dismissible in mathematics pretty much until an example of this type of problem was brought up. More explicitly, no one really cared about the complex roots of a quadratic (familiar hopefully from high school), however for a cubic, we know there is always at least one real root! More over, when there are three real roots, finding all of them absolutely requires complex arithmetic. So in short, to answer your question a root is a complex root when it is not real (so of the form a + bi with b not 0), however, the first time they were considered needed was not for themselves, but instead to find the other real roots. User-13166949966061750191 described exactly the important complex roots needed in this process.