de moivre's theorem?

Question

use de moivre's theorem to solve z^5=1

please help me, im so lost!
thanks!

we havent learnt a formula...

i have
z=cos t(theta) + i sin t
(cos t +i sin t )^5 = 1

we havent learnt a formula...

i have
z=cos t(theta) + i sin t
(cos t +i sin t )^5 = 1
= cos5t + isin5t
cos5t + isin5t = 1 +0i
cos 5t = 1
and sin 5t=0

now im stuck...

Answer 1

This is a problem involving finding the nth root. In this case, you have to find the 5 fifth roots of 1.

To do this, we first transform it into polar form.

1 = a + bi
1 = 1 + 0i

a = 1, b = 0

r = sqrt (a^2 + b^2)
r = sqrt 1
r = 1

cosθ = a/r = 1/1
θ = 0 degrees
sinθ = b/r
θ = 0 degrees

Thus:

1 = 1(cos0 + isin0)

Now use the formula

1^(1/5) [ cos ((0 + k2π)/5) + isin((0 + k2π)/5)]

k = 4, 3, 2, 1, 0. k = 0, 1 ... (n-1)


1^(1/5) [ cos ((0 + 4(2π))/5) + isin((0 + 4(2π))/5)]
1^(1/5) [ cos ((0 + 3(2π))/5) + isin((0 + 3(2π))/5)]
1^(1/5) [ cos ((0 + 2(2π))/5) + isin((0 + 2(2π))/5)]
1^(1/5) [ cos ((0 + 1(2π))/5) + isin((0 + 1(2π))/5)]
1^(1/5) [ cos ((0 + 0) + isin((0 + 0)]

You have:

cos 8π / 5 + isin 8π / 5
cos 6π / 5 + isin 6π / 5
cos 4π / 5 + isin 4π / 5
cos 2π / 5 + isin 2π / 5
1

There you go, just find the cos and sin of the angles and those are your roots.

Answer 2

you need change it into polar form so...

z^5 = 1cis0

use the de moivre's theorem:
z = 1^(1/5) cis ((0+2nπ)/5)

since it's to the power of 5 there should be 5 solutions so...

when n = 0 then z = 1 cis 0 = 1
when n = 1 then z = 1 cis (2π/5)
when n = -1 then z = 1 cis (-2π/5)
when n = 2 then z = 1 cis (4π/5)
when n = -2 then z = 1 cis (-4π/5)

Answer 3

De Moivre`s Theorem
1^(1/5) = cos (2kπ/5) + i sin (2kπ/5)
for k = 0 , 1 , 2 , 3 , 4

Roots are then:-
z1 = 1
z2 = cos (2π/5) + i sin (2π/5)
z3 = cos (4π/5) + i sin (4π/5)
z4 = cos (6π/5) + i sin (6π/5)
z5 = cos (8π/5) + i sin (8π/5)

Answer 4

1 = exp(i 2 k pi) = z^5

then

z = exp(i 2 k π / 5)

There are 5 different solutions, corresponding to k=0,1,2,3,and 4.

z1 = 1
z2 = exp(i 2 π / 5)
z3 = exp(i 4 π / 5)
z4 = exp(i 6 π / 5)
z5 = exp(i 8 π / 5)

which are the vertices of a regular pentagon inscribed in the unit circle.

Answer 5

(e^i*theta)^5=1=e^(i*2*pi)
5*theta=2*pi
theta=2*(pi/5)
z= Cos(2.pi/5)+jsin(2pi/5) where j=sqrt(-1)

Answer 6

z^5=1
1=1cis0
Let z be r cis x
(r cis x)^5=1cis0
r^5cis 5x=1cis0
r=1 and x=0
The other x's are x=72,144,216,288
The answer is cis0,cis72,cis144,cis216,cis288 and that is all!
QED