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Complex numbers and Polynomials

Terms in this set (69)

real part expression (blue)
...
imaginary part expression (imaginary)
...
shared part (purple)
...
Get sum, product, and sum of products of pairs of roots for original polynomial, e.g. {α+β = ⁻ᵇ/ₐ ; αβ = ᶜ/ₐ}. Given the roots of the new one, evaluate same summations in terms of the original equation [e.g. roots of α,β and α+β in a cubic, giving sum of roots = α+β+(α+β) = ⁻²ᵇ/ₐ ; sum of pair products = αβ + α(α+β) + β(αβ) = αβ + (αβ)² = ᶜ/ₐ + b²/a² ; product of roots = αβ(α+β) = ⁻ᶜᵇ/ₐ²]. Then just put these back into their places into the new polynomial, taking into account the symbol of their terms : e.g. x³+ (²ᵇ/ₐ)x² + (ᶜ/ₐ + b²/a²)x + (ᶜᵇ/ₐ²) ; which could be further simplified to ax³+ 2bx² + (c + b²/a)x + ᶜᵇ/ₐ
Discuss problems dealing with expressing polynomials with different roots derived from a given polynomial.
[Polynomial — Many names / terms] An expression with algebraic terms (terms of x) in different orders (ex. x²)
What is a polynomial?
Quadratic formula, completing the square, factorising, graphically
How can you solve a quadratic?
Quadratic
What is the common name for the order x²?
Cubic
What is the common name for the order x³?
Quartic
What is the common name for the order x⁴?
Quintic
What is the common name for the order x⁵?
Large coefficients or no real roots (factorising), Too rough (graphically)
What are some of the limits faced with the methods of solving a quadratic?
(±)i
What is √-1?
√(4×-1) ∴ 2i
What is √-4?
Factor theorem and comparing coefficients : write the polynomial in the form ax³ + bx² + cx + d = 0. Guess the first factor from the factors in the last term and extract the first solution from this, plug them into a formula until you get one which = 0 ; e.g. 2³-2²-(8×2)+12 = 0]. Through comparing coefficients, realise that the other term contributing to make the x² term will make the term equal to the other side [(x-2)(?) = x³-x²-8x + 12 ; (x)(bx)+(-2)(x²) = -x² ; bx² -2x² ; b-2 = -1 ; b = 1, giving a new factor (x-1), then substitute the value b and factorise to get the last factor (x+3)]
How can polynomials be solved? [x³-x²-8x+12]
Divide (mark as quotient), multiply (quotient and divisor), subtract (from dividend), bring down (quotient), repeat. In polynomial division, as the order decreases with each step, put the quotient of the first along one term, so it is above the correct order.
How to do long division — for polynomial division
The number with an equal real part but an imaginary part equal
which in magnitude and opposite in sign. It is denoted by [e.g. (x+iy) = x-iy]. The sum and product of these complex conjugates are real.
What is a complex conjugate?
It is equal either to i or -i [√(-1) or -√(-1)]
If the power of i is odd...
It is equal to 1 or -1 [√(1) or -√(1)]
If the power of i is even...
It is equal to 1
If the power of i is divisible by 4...
Where n is the degree of the highest degree term, the number of maximum turning points = 1 less than degree of highest degree term (n-1). Number of maximum roots [x-intercepts, solutions and factors] = degree of highest degree term (n).
Discuss the relationships between the number of maximum turning points and roots of a polynomial and its highest degree term
Re(x) and Im(y)
How are the real and imaginary calculations signified in a written example?
Often 'z' and 'w'
What numbers symbolise complex numbers?
It is possible for all roots to be complex
If the highest order of a polynomial is divisible by two...
Multiply numerator and denominator by its complex conjugate to give a real (rationalised) denominator.
How is it possible to simplify a fraction with a complex denominator?
A diagram that shows the complex plane of numbers, where each point is a real or complex number, the horizontal reflection (in the x — real — axis) of each point is its complex conjugate.
What is an Argand Diagram?
(x, y) where x and y are perpendicular axes on a rectangular 'Cartesian' plane
How are Cartesian coordinates written?
z = r(cosθ + isinθ) [where r is the modulus of z, and θ is the argument]
How are modulus argument polar coordinates written?
the positive square root of the sum of the squares of the real and imaginary parts of a complex number. It is written as r = |z|
What is the modulus 'r' of a complex number?
the argument of a complex number is the angle between the positive real axis and the modulus. It is a multi-valued function operating on the nonzero complex numbers. According to trigonometry, the argument of a complex number = tan⁻¹(ʸ/ₓ), where x and y are Cartesian coordinates of the point. It is important that this value makes sense from the point 0 to π or −π radians, given the quadrant the angle is in. Remember we are measuring the angle all the way around from the point zero radians to the angle of the argument, anticlockwise if in the 1st and second quadrants, or anticlockwise if in the 3rd and fourth [negative].
What is the argument of a complex number?
[multiply / divide moduli] then [add arguments]
Describe the process of multiplying and dividing complex numbers in modulus-argument form?
A half turn rotation [180°]
What transformation of the Argand diagram maps the complex number z to −z?
A reflection in the real axis
What transformation of the Argand diagram maps the complex number z to z* [z̄]?
take the modulus as the radius of the triangle, and deduce in which quadrant the point is from the argument, and take this argument from π to give the angle inside the triangle, from the radius to the real axis
How to change a complex number from modulus-argument form to rectangular form
it is equivalent to translating the point P(x,y) through (ᵃᵦ) in the argand diagram
If you add a + ib to a complex number...
It is equivalent to enlargement | a+ib | followed by an anticlockwise rotation equal to the arg(a+ib)
If you multiply a complex number by (a+ib)...
the length of the line joining w to z
|z-w| is...
the direction as an angle from w to z
arg(z-w) is...
(x+a)²+(y+b)²=r² given through completing the square, where r is the radius and (-a,-b) is the centre
What is the circle equation?
let z = x + iy , then make | (x+iy) + ai | = b then turn the modulus into an x and y value via splitting it into a real and imaginary component under a radical sign √((x)² + (y+a)²) = b , expand to give √(x² + y² + 2ay + a²) = b, square both sides so x² + y² + 2ay + a² = b² , then complete the square to form (x)² + (y+a)² = b² so the radius is b and the centre is (0, -a)
To find the loci of points where | z+ai | = b ...
substitute z for (x+iy), turn the moduli into a value of x and y under a radical symbol √((x²) + (y+a)²) = √((x²) + (y+b)²) , square the sides, expand and simplify for a value.
To find the loci of points where | z + ai | = | z + bi | ...
substitute z for (x+iy), turn the moduli into a value of x and y under a radical symbol √((x²) + (y+a)²) = c√((x²) + (y+b)²) , square the sides, expand and simplify for a value.
To find the loci of points where | z+ai | / | z+bi | = c ...
Complex numbers are such that they can have many positive and negative arguments in the argand diagram. A complex number's principal argument is its shortest — therefore in the shortest direction from zero.
What is the *principal* argument of a complex number?
x + yi where x is the real component and y is the imaaginary component
What is the form of a complex number in rectangular form?
√(Re(z))² + (Im(z))²
What is the formula for the modulus of a complex number?
r(cos(-θ) + isin(-θ)) [since cos(-θ) = cos(θ) and sin(-θ) = -sin(θ) : CHECK GRAPHS]
If a complex number is in the form r(cos(θ) - isin(θ)) what is another way this can be represented?
a circle with centre (a,0) and radius √k
The expression |z-a| = k represents a locus that resembles...
the perpendicular bisector of a and b
The expression |z-a| = |z-b| represents a locus that resembles...
the line connecting the origin and w, minus the leading part to w from the origin, where the arg(z-w) would be anticlockwise negative, and arg(z) would be anticlockwise positive hence not equal
The expression arg(z) = arg(z-w) represents a locus that resembles...
There will either be 2 distinct complex or real roots, or one repeated root, complex or real — it will always have 2
On the nature of the roots of a quadratic equation [complex or real coefficients]
There will either be 3 distinct complex or real roots, 2 distinct real roots where one is repeated, 1 root repeated 3 times, or 1 real root and 2 complex roots — it will always have 3
On the nature of the roots of a cubic equation with real coefficients...
3 distinct roots , 2 distinct roots where 1 is repeated, or 2 root repeated 3 times
On the nature of the roots of a cubic equation with complex coefficients...
A polynomial of degree n with complex (or real) coefficients will always have n roots provided we include repetitions in our counting process. This was first proven by a mathematician Gauss.
What is the fundamental theory of algebra?
First expand inside the brackets of RHS (right hand side), which will be in the form k(x-α)(x-β)(x-γ) and show that the degrees of x correspond to one another on both sides [LHS = ax³ + bx² + cx + d][RHS = k(x³-x²γ-βx²+βxγ-αx²+αxγ+αβx-αβγ)] ; i.e. kx³ = ax³ ∴ k = a {from now on use a} ||| a(-αx²-βx²-γx²) = bx² ∴ [by dividing by x²] b = -a(α+β+γ) {sum of roots is -b/a = α+β+γ} ||| cx = a(βxγ+αxγ+αβx) ∴ c = a(βγ+αγ+αβ) {sum of product of roots two at a time is c/a = βγ+αγ+αβ} ||| Finally, d = a(-αβγ) ∴ {product of all roots = -d/a} ||| Finally, d = a(-αβγ) ∴ {*product of all roots* = -*d/a* = αβγ }
Describe the process of taking coefficients of a cubic for the sum of the roots, sum of two roots and product of roots
Expand inside brackets of RHS. Make ax³ = kx³ when the product of the x terms are x³. Orders of x on each side LHS = RHS, and hence in dividing by these orders in separate equations e.g. bx² = [a(-nx² +-lx²-ox² )] to give equations for a, b, c etc. we get values inside the brackets of the sum and product of roots, and the sum of the products of two roots at a time. Divide by the constant outside the brackets andf simplify so any minus is on the top of the fraction to give the equated value for these.
Describe the process of taking coefficients of any polynomial for the sum of the roots, sum of two roots and product of roots
1) Declare z = x + iy — 2) Simplify fractions of z by multiplying complex conjugates — 3) If asked to give the locus of a second complex number set, given the locus of the other, substitute the values of u and v from Q(u,v) for example in terms of x and y from P(x,y) into P's locus
Steps to answering a question with a transformation / two different complex number sets
-b/a
After taking coefficients to prove such, how can the sum of
the roots be expressed as a fraction of the coefficients?
c/a
After taking coefficients to prove such, how can the product of
the roots of a quadratic [or sum of product of two roots in higher orders] be expressed as a fraction of the coefficients?
-d/a
After taking coefficients to prove such, how can the product of
the roots of a cubic [or sum of product of three roots in higher orders] be expressed as a fraction of the coefficients?
e/a
After taking coefficients to prove such, how can the product of
the roots of a quartic [or sum of product of four roots in higher orders] be expressed as a fraction of the coefficients?
-f/a
After taking coefficients to prove such, how can the product of
the roots of a quintic [or sum of product of five roots in higher orders] be expressed as a fraction of the coefficients?
We know that values for a, -b/a, c/a, -d/a, and e/a will be given by products and sums of certain roots, hence after we know a, it may be calculated by evaluating these fractions and putting the values for b,c,d, and e into the polynomial coefficients respectively
Given the roots of a polynomial, how might the polynomial be calculated (without expansion)?
Because the sum or product or sum of products of the roots equal to certain fractions of the coefficients of the polynomial. It's possible to calculate the value of these corresponding fractions, and by operating on these expressions before modifying to give the final expression (i.e. (α+β+γ)² = α²+β²+γ² + 2(αβ+βγ+γα)) You can hence calculate the expression by subtracting to give α²+β²+γ²
Given a polynomial equation and unknown values for the roots, how might an expression of these roots (i.e. α²+β²+γ²) be enumerated?
If you have calculated the sum of the squares of the roots, then you know that the sum of the squares of the roots of a polynomial who has all real roots is at least zero since i² = −1. If the sum of the squares is negative, then we know it has at least 2 complex roots since complex roots appear in conjugate pairs in a polynomial with real coefficients. Also, consider the shape of the polynomial function, if it is for example a continuous polynomial, where if the order is even number (e.g. ²,⁴,⁶) it has at least two real roots (goes round and back through) and if the order is odd (e.g. ¹,³,⁵,⁷) it has at least one (goes through and bends a bit before proceeding to increase)
How to deduce how many real roots a polynomial has?
Let the original polynomial = y ; When x = 0, then y = +c (from the polynomial) ; next make 2α+3, 2β+3, 2γ+3 = y [since y = 0, and these also = 0 since they are roots], hence α, β, & γ = (y−3)/2 [it is a coincidence that they are all equal in this circumstance]. Next substitute the values into the original polynomial, so [((y-3)/2)⁴ - 3(y-3)/2 + 126((y-3)/2) - 136] which simplifies to y⁴ -18y³+156y²+450y-4525
Given a polynomial equation {x⁴-3x³+12x²+126x-136=0} of roots α,β, and γ, how can another polynomial of roots 2α+3, 2β+3, 2γ+3 be constructed?
Recognise that if the polynomial has real coefficients, the complex conjugate, 2-i, is also a root, hence (x-(2+i))(x-(2-i)) is a root expanding this to give (x²-4x+5)(x²+ax+1) since 5 × 1 = 5, then by taking the coefficient for x³ terms, one would recognise that -2x³ = ax³-4x³ hence a = 2. By putting this back into the second quadratic to give (x²+2x+1) we can factorise to give the remaining roots which in this case are repeated roots of -1. So roots : 2+i, 2-i,-1
Given a complex root of a polynomial {x⁵-2x³-2x²+6x+5 = 0} is 2+i, how might one determine the remaining roots?
Factors of 28 shows f(-4) = 0, hence (x+4) is a factor of the polynomial ; Put in the form (x+4)(x²+ax+7) then by taking coefficients of x², it is shown that 4x²+ax²=2², 4+a=2 hence a=-2. Put this back into the quadratic (x²-2x+7) then put the quadratic into the quadratic formula to solve for the final roots [1 + i√6 and 1-i√6 *1-i√6*]
Given a complex root of a polynomial {x³+2x²-x+28} how would you go about deducing the roots?
multiply by denominator complex conjugate
...
substitue z as x+iy and simplify real and imaginary parts
...
It will be the square of the sum of the roots derived from the polynomial coefficients, minus 2 times the product of the roots. [E.g. in an expression ax² + bx + c, the sum of the roots will be ⁻ᵇ/ₐ, andthe product will be ᶜ/ₐ therefore the sum of the squares of the roots is (⁻ᵇ/ₐ)² −2(ᶜ/ₐ) since (⁻ᵇ/ₐ)² = r₁² + r₂² + 2(r₁r₂))
How to calculate the sum of the squares of the roots of a polynomial?
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