Consider the cubic polynomial
f(z) =z3−2z+ 4.
Given that z = 1 +i is a root of f(z), determine all the roots of f(z) in C
f(z) = z^3 - 2z + 4 , note how we type exponents
all complex roots , and irrational roots, come in conjugate pairs,
so the roots are 1+i and 1-i
sum of those is 2, product of those is also 2, so the quadratic producing
those is (z^2 - 2z + 2)
by long division that gives: (z+2)(z^2 - 2z + 2)
IF (z+2)(z^2 - 2z + 2) = 0
z = -2, 1 ± i
How do you go about getting a quadratic from the sum and product? Sorry about this I'm not great with root questions
based on the fact that if you have ax^2 + bx + c = 0,
the sum of the roots = -b/a
and the product of the roots = c/a
often this is a avery efficient way to proceed, especially if the roots are irrational or complex.
Unfortunately, this does not appear to be on the curriculum of many regions, too bad.
e.g.
(2x+1)(x-3) = 0 , or 2x^2 - 5x - 3 = 0
from the factored form, x = -1/2 and 3
sum of those roots = -1/2 + 3 = 5/2
product of roots = (-1/2)(3) = -3/2
notice that -b/a = -(-5/2) = 5/2 and c/a = -3/2 , as I stated above
Okay that really helps. Thank you!
To determine all the roots of the cubic polynomial f(z) = z^3 - 2z + 4 in the complex numbers (C), we can use the fact that if z = a + bi is a root of f(z), then its complex conjugate z* = a - bi is also a root of f(z).
Given that z = 1 + i is a root of f(z), its complex conjugate is z* = 1 - i. So we know that both z = 1 + i and z* = 1 - i are roots of f(z).
Now, let's find the remaining root by dividing f(z) by the quadratic factor (z - z1)(z - z2), where z1 = 1 + i and z2 = 1 - i.
Dividing f(z) by (z - z1)(z - z2):
f(z) / (z - z1)(z - z2) = (z^3 - 2z + 4) / (z - 1 - i)(z - 1 + i)
To simplify this expression, we can use polynomial long division or synthetic division. Let's use synthetic division:
1+i │ 1 0 -2 4
│ 1+i -2
────────────
1 1+i -1+i 2
After performing the division, the quotient is z^2 + (1+i)z - 1+i and the remainder is 2.
Now, we set the quotient equal to zero to find the remaining roots:
z^2 + (1+i)z - 1+i = 0
To solve this quadratic equation, we can use the quadratic formula:
z = (-b ± sqrt(b^2 - 4ac)) / (2a)
Plugging in the values:
z = (-(1+i) ± sqrt((1+i)^2 - 4(1)(-1+i))) / (2(1))
Expanding and simplifying:
z = (-1 - i ± sqrt(1 + 2i - 1 - 4 + 4i - 4)) / 2
z = (-1 - i ± sqrt(2i)) / 2
z = (-1 - i ± sqrt(2)(cos(π/4) + i sin(π/4))) / 2
z = (-1 - i ± sqrt(2)/√2 (cos(π/4) + i sin(π/4))) / 2
z = (-1 - i ± (√2/2)(1 + i)) / 2
Now, we can simplify this expression further:
z = (-1 - i ± (√2/2 + √2/2)i) / 2
z = (-1 - i ± (√2/2)i - (√2/2)i) / 2
z = (-1 - i ± (√2/2)i) / 2 - (√2/2)i / 2
Simplifying the fractions:
z = (-1 - i ± (√2/2)i) / 2 - (√2/2)i / 2
z = (-1 - i ± (√2/2)i) / 2 - (√2/2)i / 2
z = (-1/2 - 1/2i ± (√2/2)i/2) - (√2/2)i/2
z = (-1/2 ± (√2/2)i) - (√2/4)i
z = -1/2 ± (√2/2)i - (√2/4)i
z = -1/2 ± (2√2 - √2)/4)i
z = (-1 ± √2 - √2/4)i
Therefore, the roots of the cubic polynomial f(z) in C are:
z1 = 1 + i
z2 = 1 - i
z3 = -1 + √2 - √2/4i
z4 = -1 - √2 + √2/4i