A Nice Cubic Equation | Problem 388
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- Опубликовано: 8 окт 2024
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z³ + 3z² + 3z = i
z³ + 3z² + 3z - i = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power
Let: z = x - (b/3a) → where:
b is the coefficient for x², in our case: 3
a is the coefficient for x³, in our case: 1
z³ + 3z² + 3z - i = 0 → let: z = x - (3/3) → let: z = x - 1
(x - 1)³ + 3.(x - 1)² + 3.(x - 1) - i = 0
(x - 1)².(x - 1) + 3.(x² - 2x + 1) + 3x - 3 - i = 0
(x² - 2x + 1).(x - 1) + 3x² - 6x + 3 + 3x - 3 - i = 0
x³ - x² - 2x² + 2x + x - 1 + 3x² - 6x + 3 + 3x - 3 - i = 0
x³ - 1 - i = 0
x³ = 1 + i ← this is a complex number
The modulus is: m = √[(1)² + (1)²] = √2 → the modsulus of x is: m^(1/3) → (√2)^(1/3) = ³√2
The argument is β sush as: tan(β) = 1/1 = 1 → β = π/4 → the argument of x is: β/3 → (π/4)/3 = π/12
So you can see that the first root of x³ is:
x1 = ³√2.[cos(π/12) + i.sin(π/12)] → to get the second root, you add 2π/3
x2 = ³√2.[cos{(π/12) + (2π/3)} + i.sin{(π/12) + (2π/3)}] → and to get the third one, you add 2π/3 once again
x3 = ³√2.[cos{(π/12) + (2π/3) + (2π/3)} + i.sin{(π/12) + (2π/3) + (2π/3)}]
It gives:
x1 = ³√2.[cos(π/12) + i.sin(π/12)]
x2 = ³√2.[cos(3π/4) + i.sin(3π/4)]
x3 = ³√2.[cos(17π/12) + i.sin(17π/12)]
Recall: z = x - 1
z1 = ³√2.[cos(π/12) + i.sin(π/12)] - 1
z2 = ³√2.[cos(3π/4) + i.sin(3π/4)] - 1
z3 = ³√2.[cos(17π/12) + i.sin(17π/12)] - 1
To go further:
cos(3π/4) = - (√2)/2
sin(3π/4) = (√2)/2
cos(17π/12) = cos[(12π + 5π)/12] = cos[(12π/12) + (5π/12)] = cos[π + (5π/12)] = - cos(5π/12)
sin(17π/12) = sin[(12π + 5π)/12] = sin[(12π/12) + (5π/12)] = sin[π + (5π/12)] = - sin(5π/12)
After simplification:
z1 = ³√2.[cos(π/12) + i.sin(π/12)] - 1
z2 = [³√2 * (√2)/2 * (1 + i)] - 1
z3 = - ³√2.[cos(5π/12) + i.sin(5π/12)] - 1
I used the second method.
We always know that A^3=B^3
A^3-B^3=0
(A-B)(A^2+AB+B^2)=0
We should find the solutions from A^2+AB+B^2=0
In this case: Z^2+Z*(1+i)^(1/3)+(1+i)^(2/3)=0
Why not add 1 to both sides then you got (Z+1)^3=I+1 and after getting left side into polar form it is very easy to get the solution
(Z+1)^3=√2(cos(π/4)+I*sin(π/4))
Z+1=2^(1/6)(cos(π/12)+i sin(π/12)) , also with arguments 5π/12 and 3π/4. To have all 3 solutions
(z + 3)³ = 1 + I
So z = - 3 + (1 + I)¹ᐟ³
it's not the right formula
@@luunguyen193
Right!