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# complex number question watch

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1. hmm, a bit stumped here

solve the equation: z^5 + 16 = 0

anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
2. (Original post by kimoni)
hmm, a bit stumped here

solve the equation: z^5 + 16 = 0

anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
As

(r (cos a + i sin a))^n = r^n (cos(na)+isin(na))

we can see that if

z^5 = -16

we need z to have a modulus that is the fifth root of |-16| which is 2^(4/5)

and will need one fifth of the argument which is pi (or 3pi or 5pi ...)

So five roots are

z = 2^(4/5) (cos(k pi/5) + i sin(k pi/5))

for k = 1,3,5,7,9

say.
3. (Original post by kimoni)
hmm, a bit stumped here

solve the equation: z^5 + 16 = 0

anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
You can do these sorts of things by thinking of circles and equally spaced points/lines on them to represent arguments. I get 16^(1/5).e(i.pi/5 + 2npi/5) ect. as solutions.

Ben
4. ok I think i'm beginning to see

how did you know k was 1,3,5,7,9 though?
5. (Original post by kimoni)
hmm, a bit stumped here

solve the equation: z^5 + 16 = 0

anyone know how to do it? I think De Moivre's theorem is supposed to be used... rep to anyone who can do it
just curious, is this a question on its own or was there a prelim bit?
6. (Original post by El Stevo)
just curious, is this a question on its own or was there a prelim bit?
just on it's own, but there are other qs like it
7. (Original post by kimoni)
ok I think i'm beginning to see

how did you know k was 1,3,5,7,9 though?
Essentially, you've got the modulus and you're just trying to get the complex part to give you the -1 which you require to satisfy the equation. To get this, using e^i.theta, you can use theta = pi and, beacause of the periodicity of trig functions, you can use every angle 2pi larger. Then you just divide your angles by 5 to satisfy the question (as you take the 5th root, which is effectively dividing the exponential power by 5).

Ben
8. (Original post by kimoni)
ok I think i'm beginning to see

how did you know k was 1,3,5,7,9 though?
The argument of -16 is an odd multiple of pi.

z = 2^(4/5) (cos a + i sin a)

where a = k pi/5 and k is odd.

But there aren't infinite roots, only five. So if you start counting through odd k then k=1,3,5,7,9 all give differents roots, but k=11 gives the same root as k=1, and k =13 the same as k=3, etc. This is because sin and cos have period 2pi.
9. ok, thanks very much chaps

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