The theorem is stated in two steps as follows:
(i) (cos +i sin )n = cos n
+i sin n , if n is an integer (positive, zero or negative)
(ii) cos n +i sin n is one of the
values of (cos + i sin )n
if n is a non integral rational number.
Let z = r cos = r cos ( +2 m
), where m I
z1/n = r1/n
Note that r1/n is a real, positive number. Putting m = 0, 1, 2, ...
n -1 we get n distinct roots of z, and then the values of
start repeating.
Since 1 = 1 +0 i = cis 0 = cis (0 +2m) = cis 2m,
m I,
11/3 = (cis 2m)1/3 = cis 2m/3
where m = 0, 1, 2
= cis 0, cis 2/3, cis 4/3
= 1, (-1 +3 i)/2, (-1 -3 i)/2
Putting w = (-1 +3 1)/2
= (1 -3 -23 i)/4 = (-1 -3 i)/2
Hence three cube roots of unity are 1, w, w² where w = (-1 +3 i)/2
If z = (13 -5 i)/(4 -9 i), find z6.
Given z = (13 -5 i)/(4 -9 i) =[(13 -5 i)(4 +9 i)]/[(4 -9i)(4 +9i)]
= (97 +97 i)/97
= 1 + i/4 = 2 [cos(/4) + i sin (/4)]
Using Demoivre theorem,
z6 =
= 8(cos 3/2 +i sin 3/2) = 8[0 +i(-1)]
= -8i
Convert the following number to standard form x +iy
(cos +i sin )7
(cos -i sin )³
(cos +i sin )7 (cos
-i sin )³ = (cis )7
(cis (- ))³
= (cis )7 ((cis
)-1)³ = (cis )7
(cis )-3
= (cis )4 =
cis 4 = cos 4 +i sin 4 .
Given that one of the roots of x4 -2 x³ +3 x² -2 x +2 = 0 is 1 +i, find the other three roots.
Since all the coefficients of the given equation are real numbers, the
roots, if complex, occur in conjugate pairs. Thus two roots are 1 +i, 1 -i.
Hence x -(1 +i), x -(1 -i) are two factors of given equation,
i.e. (x -(1 +i))(x -(1 -i)) = x² -2 x +2 is a factor of given equation.
Now it is easy to see that x4 -2 x³ +3 x² -2 x +2 = (x² -2x +2)(x²
+1).
Also we know that x² +1 = (x -i)(x +i).
Thus the four roots of the given equation are ±i, 1 ±i.
Find the fourth roots of unity.
1 = 1 +0 i = cis 0
= cis (0 +2 m) = cis 2m,
where m I,
11/4 = (cis 2 m )1/4 = cis 2m/4,
where m = 0, 1, 2, 3
= cis 0, cis /2, cis , cis 3/2
= 1, i, -1, -i.
Hence the fourth roots of unity are ±1, ±i.
Show that
(i) (1 +w -w²)6 = 64.
(ii) (1 -w +w²)7 +(1 +w -w²)7 = 128.
(iii) (2 -w)(2 -w²)(2 -w10)(2 -w11) = 49.
(i) (1 +w -w²)6 = (-w² -w²)6
[using 1 +w +w² = 0]
= (-2w²)6 = 26
w12 = 64(w³)4 = 64 [using w³ = 1]
(ii) (1 -w +w²)7 +(1 +w -w²)7 = (-w -w)7 +(-w² -w²)7
= -27. w7 -27 w14 = -27[(w³)².
w +(w³)4. w²]
= -27. (w +w²) = -27. (-1) = 128.
(iii) (2 -w)(2 -w²)(2 -w10)(2 -w11)
= [4 -2(w +w²) +w³] [4 -2w10(1 +w) +w21]
= [4 -2(-1) +1] [4 -2w (-w²) +1]
(why?)
= (4 +2 +1)(4 +2. 1 +1) = 7. 7 = 49.