Example 2.28
If z = (cosθ + i sinθ ) , show that zn + 1/ zn = 2 cos nθ and zn – [1/ zn] = 2i sin nθ .
Solution
Let z = (cosθ + i sinθ ) .
By de Moivre’s theorem ,
zn = (cosθ + i sinθ )n = cos nθ + i sin nθ
Example 2.29
Similarly,
Solution
Example 2.30
Solution
Example 2.31
Simplify
(i) (1+ i)18
(ii) (-√3 + 3i)31 .
Solution
(i) (1+ i)18
Let 1+ i = r (cosθ + i sinθ ) . Then, we get
(ii) (-√3 + 3i)31 .
Let -√3 + 3i = r (cosθ + i sinθ ) . Then, we get
Raising power 31 on both sides,
Example 2.32
Find the cube roots of unity.
Solution
We have to find 11/3 . Let z = 11/3 then z3 = 1.
In polar form, the equation z3 = 1 can be written as
z3 = cos(0 + 2kπ) + i sin(0 + 2kπ) = ei2kπ , k = 0, 1, 2,...
Example 2.33
Find the fourth roots of unity.
Solution
We have to find 11/4. Let z =11/4 . Then z4 = 1 .
In polar form, the equation z4 = 1 can be written as
z4 = cos (0 + 2kπ ) + i sin (0 + 2kπ ) = ei2kπ , k = 0, 1, 2,...
Note
(i) In this chapter the letter ω is used for nth roots of unity. Therefore the value of ω is depending on n as shown in following table.
(ii) The complex number zeiθ is a rotation of z by θ radians in the counter clockwise direction about the origin.
Example 2.34
Solve the equation z3 + 8i = 0 , where z ∈ C.
Let z3 + 8i = 0 . Then, we get
z3 = -8i
Find all cube roots of √3 + i
Solution
We have to find (√3 + i)1/3. Let z3 = √3 + i = r (cosθ + i sinθ )
Suppose z1 , z2, and z3 are the vertices of an equilateral triangle inscribed in the circle |z| = 2. If z1 = 1+ i√3 , then find z2 and z3.
|z| = 2 represents the circle with centre (0, 0) and radius 2.
Let A, B, and C be the vertices of the given triangle. Since the vertices z1 , z2 , and z3 form an equilateral triangle inscribed in the circle |z| = 2 , the sides of this triangle AB, BC, and CA subtend 2Ï€/3 radians (120 degree) at the origin (circumcenter of the triangle).
(The complex number z eiθ is a rotation of z by θ radians in the counter clockwise direction about the origin.)
Therefore, we can obtain z2 and z3 by the rotation of z1 by 2π/3 and 4 π/3 respectively.
Given that
Therefore, z2 = -2, and z3 = 1- i√3.
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