Solve the equation. Express your answer in trigonometric form.
x^5-11=0
I know the answer is:
11^1/5(cos α + i sin α) for α = 0°, 72°, 144°, 216°, 288°.
But I don't know where to even start working on this problem.
Huh? You have all the answers!
If you let r = 11^(1/5), the numbers are
r cis 0°
r cis 72°
...
The n * (cosθ + i sinθ) is the trigonometric or polar form.
a + bi is the complex or rectangular form.
Hmmm. I see that you don't know how to arrive at the answers. Well, if z = r cisθ, then
z^n = r^n * cis(nθ)
Since x^5-11=0 has 5 solutions, they must all be solutions to
z = 11^(1/5)
It appears that there is only one solution, but that is because there is only one real solution. There must also be 4 complex solutions, and if plotted, they all lie on the circle with radius 11^(1/5).
We want all the distinct points which lie on this circle, such that
z^5 = 11
11 = 11 cis 0°
But, 11 is also
11 cis 360°
11 cis 720°
...
We want solutions where z^5 puts us back on the real axis. If we go around 1,2,3,4 times, we wind up back where we started.
So, if z = 11^(1/5) cis72°,
z^5 = 11 cis 360°, so that is our 2nd root.
Taking increments of 72°, we get all the points on the circle which wind up back on the real axis if the angle is multiplied by 5.
Don't know whether that helps. If you can't follow the discussion in your textbook, I'm not sure I can make it any clearer. Do a web search on complex roots and see what hits you get.
Ok i ll do research on it. Thank you very much for your help.
To solve the equation x^5 - 11 = 0 and express the answer in trigonometric form, we can start by rewriting it as x^5 = 11.
Step 1: Find the fifth root of 11
To find the fifth root of 11, we can use the trigonometric form of complex numbers. We know that any complex number can be expressed as a magnitude (r) and an angle (θ) from the positive real axis in the complex plane. Therefore, we can write 11 as 11(cos(0) + i sin(0)), where cos(0) = 1 and sin(0) = 0.
Step 2: Express the fifth root of 11 in trigonometric form
To find the fifth root of 11, we will raise it to the power of 1/5. Using De Moivre's theorem, we can express the fifth root of 11 as:
(11)^1/5 = [11(cos(0) + i sin(0))]^1/5
Using the properties of exponents, we can simplify as follows:
[11(cos(0) + i sin(0))]^1/5 = 11^1/5(cos(0/5) + i sin(0/5))
Simplifying further, we get:
11^1/5(cos(0) + i sin(0)) = 11^1/5(cos(0°) + i sin(0°))
So, the fifth root of 11 in its trigonometric form is 11^1/5(cos(0°) + i sin(0°)).
Step 3: Identify the values of α
Since we need to express the answer in trigonometric form for α = 0°, 72°, 144°, 216°, and 288°, we can plug those values into the equation to find the corresponding answers.
For α = 0°:
11^1/5(cos(0°) + i sin(0°)) = 11^1/5(1 + i * 0) = 11^1/5
For α = 72°:
11^1/5(cos(72°) + i sin(72°)) = 11^1/5(cos(72°) + i sin(72°))
For α = 144°:
11^1/5(cos(144°) + i sin(144°)) = 11^1/5(cos(144°) + i sin(144°))
For α = 216°:
11^1/5(cos(216°) + i sin(216°)) = 11^1/5(cos(216°) + i sin(216°))
For α = 288°:
11^1/5(cos(288°) + i sin(288°)) = 11^1/5(cos(288°) + i sin(288°))
So, the solutions to the equation x^5 - 11 = 0 expressed in trigonometric form for α = 0°, 72°, 144°, 216°, and 288° are:
11^1/5 for α = 0°
11^1/5(cos(72°) + i sin(72°)) for α = 72°
11^1/5(cos(144°) + i sin(144°)) for α = 144°
11^1/5(cos(216°) + i sin(216°)) for α = 216°
11^1/5(cos(288°) + i sin(288°)) for α = 288°