Find all solutions of the equation. Leave answers in trigonometric form.
x^5-1024=0
I got 4(cos tehta + i sin tehta), tehta = 0, 2pi/5, 4pi/5, 6pi/5, 8pi/5
is this right
correct!
(except that's theta, not tehta)
To find the solutions of the equation x^5 - 1024 = 0 and express them in trigonometric form, we can use the fifth roots of unity.
First, let's rewrite the equation as x^5 = 1024. We can then express 1024 in polar form as 1024 = 1024(cos(0) + i sin(0)).
Now, we need to find the fifth roots of 1024. Since x^5 = 1024, we have x = ∛(1024) * ∛(cos(0) + i sin(0)), where ∛(1024) is the fifth root of 1024.
To find the fifth root of 1024, we can rewrite 1024 in polar form. 1024 = 1024(cos(0) + i sin(0)) = 1024(cos(2πk) + i sin(2πk)), where k is an integer.
Taking the fifth root, we have ∛(1024) = ∛(1024(cos(2πk) + i sin(2πk))) = 2(cos(2πk/5) + i sin(2πk/5)), where k ranges from 0 to 4 since we need to find all five roots.
Now, substituting this into x = ∛(1024) * ∛(cos(0) + i sin(0)), we get
x = 2(cos(2πk/5) + i sin(2πk/5)) * (cos(0) + i sin(0))
Simplifying, we have
x = 2(cos(2πk/5) + i sin(2πk/5))
Now, calculating for each value of k from 0 to 4, we have:
For k = 0: x = 2(cos(0) + i sin(0)) = 2(1 + i * 0) = 2
For k = 1: x = 2(cos(2π/5) + i sin(2π/5))
For k = 2: x = 2(cos(4π/5) + i sin(4π/5))
For k = 3: x = 2(cos(6π/5) + i sin(6π/5))
For k = 4: x = 2(cos(8π/5) + i sin(8π/5))
Therefore, the solutions to the equation x^5 - 1024 = 0, in trigonometric form, are:
x = 2, 2(cos(2π/5) + i sin(2π/5)), 2(cos(4π/5) + i sin(4π/5)), 2(cos(6π/5) + i sin(6π/5)), 2(cos(8π/5) + i sin(8π/5)).