Need help in a question of DeMoivre's theorem.Trigonometry?

x^4 * (x^3 + 1) + 1 * (x^3 + 1) = 0

(x^4 + 1) * (x^3 + 1) = 0

x^4 + 1 = 0

x^4 = -1

x = (-1)^(1/4)

x^3 + 1 = 0

x^3 = -1

x = (-1)^(1/3)

So our solutions, so far, are x = (-1)^(1/4) and x = (-1)^(1/3)

Now we can start using DeMoivre's Theorem. When I use the letter k, it stands for any integer.

(-1)^(1/4) =>

(-1 + 0i)^(1/4) =>

(cos(pi + 2pi * k) + i * sin(pi + 2pi * k))^(1/4)

DeMoivre's Theorem tells us that (cos(t) + i * sin(t))^n is equal to cos(n * t) + i * sin(n * t). This is why it's important to include all coterminal angles (which I did by adding 2pi * k), so as to not miss any possible answers.

(cos(pi * (1 + 2k)) + i * sin(pi * (1 + 2k)))^(1/4)

cos((pi/4) * (1 + 2k)) + i * sin((pi/4) * (1 + 2k))

We know that cosine and sine repeat every 2pi radians, so let's set up some bounds for k

0 </= (pi/4) * (1 + 2k) </= 2pi

0 * (4/pi) </= 1 + 2k </= 2pi * (4/pi)

0 </= 1 + 2k </= 8

-1 </= 2k </= 7

-1/2 </= k </= 7/2

k is an integer, so the only valid values for k are 0 , 1 , 2 , and 3

cos((pi/4) * (1 + 2k)) + i * sin((pi/4) * (1 + 2k))

cos((pi/4) * (1 + 2 * 0)) + i * sin((pi/4) * (1 + 2 * 0))

cos((pi/4) * (1 + 2 * 1)) + i * sin((pi/4) * (1 + 2 * 1))

cos((pi/4) * (1 + 2 * 2)) + i * sin((pi/4) * (1 + 2 * 2))

cos((pi/4) * (1 + 2 * 3)) + i * sin((pi/4) * (1 + 2 * 3))

cos(pi/4) + i * sin(pi/4) ; cos(3pi/4) + i * sin(3pi/4) ; cos(5pi/4) + i * sin(5pi/4) ; cos(7pi/4) + i * sin(7pi/4)

+/- (sqrt(2)/2) * (1 +/- i)

x = (-1)^(1/3)

x = (-1 + 0i)^(1/3)

x = (cos(pi + 2pi * k) + i * sin(pi + 2pi * k))^(1/3)

x = (cos(pi * (1 + 2k)) + i * sin(pi * (1 + 2k)))^(1/3)

x = cos((pi/3) * (1 + 2k)) + i * sin((pi/3) * (1 + 2k))

0 </= (pi/3) * (1 + 2k) </= 2pi

0 </= 1 + 2k </= 6

-1 </= 2k </= 5

-1/2 </= k </= 5/2

k = 0 , 1 , 2

x = cos((pi/3) * (1 + 2k)) + i * sin((pi/3) * (1 + 2k))

cos(pi/3) + i * sin(pi/3) , cos(3pi/3) + i * sin(3pi/3) , cos(5pi/3) + i * sin(5pi/3)

1/2 + i * sqrt(3)/2 ; -1 ; 1/2 - i * sqrt(3)/2

x =>

(sqrt(2)/2) * (1 + i)

(sqrt(2)/2) * (1 - i)

(-sqrt(2)/2) * (1 + i)

(-sqrt(2)/2) * (1 - i)

(1/2) * (1 + i * sqrt(3))

(1/2) * (1 - i * sqrt(3))

-1

Now, in case you're wondering how I can determine that -1 + 0i is equal to cos(pi + 2pi * k) + i * sin(pi + 2pi * k)

a + bi = R * (cos(t) + i * sin(t))

a = R * cos(t)

bi = R * sin(t) * i

a = R * cos(t)

b = R * sin(t)

a^2 + b^2 = R^2 * cos(t)^2 + R^2 * sin(t)^2

a^2 + b^2 = R^2 * (cos(t)^2 + sin(t)^2)

a^2 + b^2 = R^2 * 1

a^2 + b^2 = R^2

In our case, a = -1 and b = 0

(-1)^2 + 0^2 = R^2

1 + 0 = R^2

1 = R^2

-1 , 1 = R

We can go ahead and leave out -1 = R. Thanks to DeMoivre's Theorem and the algebraic completeness of complex numbers, solving for cases of R = -1 will just give us duplicate answers of solving for R = 1 (we'll just move clockwise around the coordinate plane, as opposed to moving counter-clockwise). You can go ahead and work it as R = -1 and see for yourself, if you'd like.

a = R * cos(t)

-1 = 1 * cos(t)

-1 = cos(t)

t = arccos(-1) = ... , -5pi , -3pi , -pi , pi , 3pi , 5pi , ....

b = R * sin(t)

0 = 1 * sin(t)

0 = sin(t)

t = ... , -3pi , -2pi , -pi , 0 , pi , 2pi , 3pi , .....

t must be the common elements of both sets. In this case, t = pi + 2pi * k, where k is an integer.

The rest is just plug-and-chug.

i) Factorizing, (x³ + 1)*(x⁴ + 1) = 0

ii) So, either x³ = -1 or x⁴ = -1

iii) Consider x³ = -1:

x³ = cos(π) + i*sin(π) = cos(2kπ + π) + i*sin(2kπ + π), where k is an integer.

So, x = [cos(2kπ + π) + i*sin(2kπ + π)]^(1/3)

Applying De-moivre's theorem,

x = [cos{(2kπ + π)/3} + i*sin{(2kπ + π)/3}]

At k = 0, 1st root = cos(π/3) + i*sin(π/3) = (1 + i*√3)/2

At k = 1, 2nd root = cos(π) + i*sin(π) = -1

At k = 2, 3rd root = cos(5π/3) + i*sin(5π/3) = (1 - i*√3)/2

iv) Let us now consider, x⁴ = -1:

As in previous, x = [cos(2kπ + π) + i*sin(2kπ + π)]^(1/4)

==> x = [cos{(2kπ + π)/4} + i*sin{(2kπ + π)/4}]

Rest you may evaluate, by putting k = 0, 1, 2 and 3 as in previous.