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July 23, 2014

July 23, 2014

Posted by **hello** on Monday, July 1, 2013 at 11:02pm.

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.

- Trignometry -
**Steve**, Monday, July 1, 2013 at 11:57pmHuh? 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.

- Trignometry -
**hello**, Tuesday, July 2, 2013 at 12:11amOk i ll do research on it. Thank you very much for your help.

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