by using the substitution w = z^3, find all the solutions to z^6 - 8z^3 +25 = 0 in complex numbers, and describe them in polar form, using @(theta) to denote the angle satisfying tan@ = 3/4 ( note simply leave @ as it is, don't calculate it).
i got up to z^3 = 4+3i and 4-3i then got stuck !
4 + 3i = 5 Exp[i theta]
The equation z^3 = Q for real positive Q has three solutions:
z = cuberoot[Q] Exp[2 pi n i/3]
for n = 0, 1 and 2, because
Exp[2 pi n i/3]^3 = Exp[2 pi n i] = 1
So, in this case you find:
z = 5^(1/3) Exp[i theta/3 + 2 pi n i/3 ]
To solve the equation z^6 - 8z^3 + 25 = 0 using the substitution w = z^3, you correctly obtained z^3 = 4 + 3i and z^3 = 4 - 3i. Now, let's find the solutions for z.
1. For z^3 = 4 + 3i:
To express 4 + 3i in polar form, we need to find its magnitude and angle.
Magnitude (r):
The magnitude of a complex number z = a + bi is given by |z| = sqrt(a^2 + b^2).
Here, a = 4 and b = 3, so |4 + 3i| = sqrt(4^2 + 3^2) = 5.
Angle (theta):
The angle theta is given by tan(theta) = b/a.
Here, b = 3 and a = 4, so tan(theta) = 3/4.
So, for z^3 = 4 + 3i, z = 5^(1/3) Exp[i (theta/3 + (2pi*n)/3)].
2. For z^3 = 4 - 3i:
Again, we need to find the magnitude and angle of 4 - 3i.
Magnitude (r):
|4 - 3i| = sqrt(4^2 + (-3)^2) = 5.
Angle (theta):
tan(theta) = -3/4.
So, for z^3 = 4 - 3i, z = 5^(1/3) Exp[i (theta/3 + (2pi*n)/3)].
By substituting these values of magnitude and angle into the polar form, we get:
For z^3 = 4 + 3i:
z = 5^(1/3) Exp[i (theta/3 + (2pi*n)/3)].
For z^3 = 4 - 3i:
z = 5^(1/3) Exp[i (theta/3 + (2pi*n)/3)].
Note that we don't need to calculate the value of theta, we can leave it as it is denoted by the symbol @. The value of theta can be found using tan@=3/4, but we are instructed to keep it in the symbolic form.