MATHEMATICS

ကိုယ်တိုင်လည်းလေ့လာဆည်းပူးရင်း၊ လိုအပ်သူများအတွက် အချက်အလက်တချို့ကို ရေးတင်ထားပေးရင်းနဲ့ အကြောင်းအရာတချို့ကို တင်ဆက်ပေးပါမယ်။ နိုင်ငံတကာမှာ အခြေခံပညာနဲ့ အဆင့်မြင့်ပညာရေး သင်ရိုးစာအုပ်ပါ အကြောင်းအရာတွေနဲ့ အထောက်အကူပြုစာတွေကို အွန်လိုင်းပေါ် အဆင်သင့်တင်ပေးနေကြသူများရှိနေတဲ့နည်းတူ မြန်မာနိုင်ငံမှာလည်း ရှိစေချင်တာကြောင့် အချိန်ကုန်၊ ငွေကုန်ကြေးကျခံပြီးလုပ်ထားတဲ့ ဝက်ဘ်ဆိုက်ပိစိကွေးလေးပါ။

အကြောင်းအရာ
1.1 Pure Imaginary Unit i
1.2 Complex Number (x,y)
1.3 Operations and Complex Numbers
1.4 Trigonometric Form
1.5 Roots of Complex Numbers
အခြားဖတ်စရာများ

GRADE 12

ဖုန်းဖြင့်ကြည့်ချိန် အပြည့်မမြင်ရလျှင် ဖုန်းကို အလျားလိုက်လှဲထားပြီးကြည့်ရနိုင်ပါသည်။

Chapter 1

Complex Numbers

In this chapter we introduce a new number system which is the extension of the real number system.

1.1 Pure Imaginary Unit i

Since we have known that x2 ≥ 0 for every real number x, the equation x2 = -4 has no real solution. But if there is a pure imaginary unit ⅈ such that
i2 = -1
with acceptance of the usual operations on real numbers and ⅈ such as
(2ⅈ)2 = 4ⅈ2 = 4(-1) = -4 and (-2ⅈ)2 = 4ⅈ2 = 4(-1) = -4
then x2 = -4 has two solutions 2ⅈ and -2ⅈ.
If we try to solve x2 = -n, for n > 0, like x2 = -4, then √ n ⅈ and -√ n ⅈ are solutions because
(±√ n ⅈ)2 = nⅈ2 = n(-1) = -n

Example 1.

Solve x2 + 2x + 5 = 0.
Solution

x2 + 2x + 5 = 0
x2 + 2x + 1 = -4
(x + 1)2 = -4
= 4 (-1)
x + 1 = ±√ 4 √ i2
x + 1 = ±√ 4 i
x = -1 ±2i

So there are two solutions x = -1 + 2i and x = -1 - 2i.

Example 2.
Solve x2 + 2x + 3 = 0 and check your answers.

Solution
x2 + 2x + 3 = 0
x2 + 2x + 1 = -2
(x + 1)2 = -2
= 2 (-1)
= 2 i2
x + 1 = ±√ 2 √ i2
x + 1 = ± √ 2 i
x = -1 ± √ 2 i

So two solutions are x = -1 + √ 2 i and x = -1 - √ 2 i

For x = -1 + √ 2 i,
x2 + 2x + 3 = (-1 + √ 2 i)2 + 2(-1 + √ 2 i) + 3
= 1 - 2√ 2 i + 2i2 - 2 + 2√ 2 i + 3
= 1 - 2 √ 2 i + 2(-1) - 2 + 2 √ 2 i + 3
= 1 - 2√ 2 i - 2 -2 + 2√ 2 i + 3
= 0

For x = -1 - √ 2 i,
x2 + 2x + 3 = (-1 - √ 2 i)2 + 2(-1 - √ 2 i) + 3
= 1 + 2√ 2 i + 2i2 - 2 - 2√ 2 i + 3
= 1 + 2√ 2 i - 2 -2 - 2√ 2 i + 3
= 0

Exercise 1.1

  1. Solve the following equations.

    2. (a) x2 - 6x + 10 = 0
    (b) -2x2 + 4x - 3 = 0
    (c) 5x2 - 2x + 1 = 0
    (d) 3x2 + 7x + 5 = 0

  2. Solve the following equations and check your answers.

    3. (a) x2 - 2x + 4 = 0
    (b) x2 - 4x + 5 = 0

  3. Find the value of in for every positive integer n, where i2 = -1, i3 = i2i , i4 = i2i2, etc.

1.2 Complex Number (x, y)

As we have just seen in section 1.1, there are numbers like x1 + y1i and x2 + y2i such that
(x1 + y1 i) + (x2 + y2 i) = (x1 + x2) + (y1 + y2)i and
(x1 + y1i) ( x2 + y2i) = x1x2 + (x1y2 + y1x2)i + y1y2i2
= x1x2 + (x1y2 + y1x2)i + y1y2(-1)
= (x1x2 - y1y2) + (x1y2 + y1x2)i

for real numbers x1, x2, y1, y2, so we define the number x + yi as follows.

A complex number is an ordered pair (x, y) of real numbers with equality and operations— sum and product— of two complex numbers (x1, y1), (x2, y2) are defined as follows:
Equality (x1, y1) = (x2, y2) if and only if x1 = x2 and y1 = y2
Sum (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2)
Product (x1, y1)(x2, y2) = (x1x2 - y1 y2, x1 y2 + y1 x2)


Then we have
(x, 0) + (y, 0) = (x + y, 0) and (x, 0)(y, 0) = (xy, 0)

From these facts, all real numbers can be considered as complex numbers as shown in the following table.
Real Numbers     Complex Numbers
x (x, 0)
y (y, 0)
x + y (x, 0) + (y, 0)
xy (x, 0)(y, 0)

Since (0, 1)(0, 1) = (0 - 1, 0 + 1) = (-1, 0), let us denote
i = (0, 1)
and with the convention i2 = ii, i3 = i2i, etc., we have
i2 = (0, 1)(0, 1) = (0 - 1, 0 + 0) = (-1, 0) = -1

And we have
x + yi = (x, 0) + (y, 0)(0, 1)
= (x, 0) + (0 - 0, y + 0)
= (x, 0) + (0, y)
x + yi = (x, y)

Note that sum and product of complex numbers satisfy commutative, associative and distributive properties.

Example 3.
Compute (-2, 3)(1, -2) + (1, 1)(0, 1).

Solution
Method 1
(-2, 3)(1, -2) + (1, 1)(0, 1) = (-2 - (-6), 4 + 3) + (0-1, 1+0)
= (4, 7) + (-1, 1) = (3, 8)

Method 2
(-2, 3)(1, -2) + (1, 1)(0, 1) = (-2 + 3i)(1-2i) + (1 + i)i
= (-2 + 4i + 3i - 6i2) + (i + i2)
= (-2 + 7i - 6(-1)) + (i - 1)
= (4 + 7i) + (-1 + i)
= (3 + 8i)
= (3, 8)

Exercise 1.2
  1. Compute:
    2. (a) (2, 0)(3, 5) + (3, -2)(0, 1)
    (b) (2, -5)(-1, 0) + (1, 0)(5, 1)
    (c) (-3, -2)(-2, -3) + (-2, -3)(-3, -2)
    (d) (1, 0)(0, 1) + (0, 1)(1, 0)

  2. Compute:
    3. (a) (3 + 2i)(3 - 2i) + (-5 + 7i)(-1 - i)
    (b) (-1 + i)(1 - i) + (2 + 3i)
    (c) (1 + i)(1 - i) + (-2 + i)(-2 + i)
    (d) (3 + 2i) + (7 - i)(-3 + 3i)


1.3 Operations on Complex Numbers

Sum and product of complex numbers were defined in Section 1.2. Subtraction of complex numbers is defined as in real numbers as follows:

(x1+ y1i) - (x2 + y2i) = (x1 + y1i) + (-x2 - y2 i)


Therefore sum, product and subtraction of complex numbers can be performed as in the real numbers, except only that i2 = -1. But the division of complex numbers is a little different from the division of real numbers. We need some notations to define the division.
From now on let us denote a complex number by z, so that

z = (x + y i) = (x, y)

$$ \text{the \textbf{conjugate}}\ \bar{z} \ \text{of a complex number}\ z = x + yi $$ $$= (x, y) \text{is defined by} $$ $$ \bar{z} = x - yi = (x, -y) $$ Then we have $$ z\bar{z} = (x + yi)(x - yi) = x^2 + y^2 $$ Let z1 = x1 + y1i and z2 + y2i.
We will calculate z1⁄z2, (z2 ≠ 0), as follows:
$$ \frac{z_1}{z_2} = \frac{z_1}{z_2} \frac{\bar{z}_2}{\bar{z}_2} $$ $$ = \frac{x_1 + y_1i}{x_2 + y_2i} \frac{x_2 - y_2i}{x_2 - y_2i} $$ $$ = \frac{x_1x_2 + y_1y_2 (y_1x_2 - x_1y_2)i}{x_2 ^2 + y_2^2 } $$ $$ = \frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} + \frac{y_1x_2 - x_1y_2}{x_2^2 + y_2^2}i $$
Example 4.
Calculate 2 + 3i ⁄3 + i

Solution
2 + 3i ⁄3 + i = 2 + 3i ⁄3 + i 3 - i ⁄3 - i
= 6 + 3 + (9 - 2)i ⁄9 + 1
= 9 + 7i⁄10
= 9 ⁄10 + 7 ⁄10 i
Now we calculate 1 ⁄z for z = x + yi, z ≠ 0,
1 ⁄z = 1 ⁄x + yi x - yi ⁄x - yi
= x - yi ⁄x2 + y2
= x ⁄x2 + y2 - y ⁄x2 + y2i

And we can check that
z (1 ⁄z) = (x + yi) (x ⁄x2 + y2 - y ⁄x2 + y2 i)
= x2 + y2 ⁄x2 + y2 + yx - xy ⁄x2 + y2 i
= 1

So, for a non-zero complex number z, 1 ⁄z is the multiplicative inverse of z and is denoted by z-1. From these facts, division of complex numbers is defined as
z1⁄z2 = z1z2-1
but we usually calculate z1 ⁄z2 as in Example 4.

Exercise 1.3
1. Let z1 = -2 + 3i, z2 = 5 + 2i. Compute:
$$ \text{(a)}\ z_1^2 - 2z_1 + 1 $$ $$ \text{(b)}\ 3z_2^2 + 2z_2 - 1 $$ $$ \text{(c)}\ z_1\bar{z}_2 + z_2\bar{z}_1 $$ $$ \text{(d)}\ \frac{1}{z_1} $$ $$ \text{(e)}\ \frac{1}{z_2} $$ $$ \text{(f)}\ \frac{1}{z_1z_2} $$ $$ \text{(g)}\ \frac{z_1}{z_2} $$ $$ \text{(h)}\ \frac{\bar{z}_1}{\bar{z}_2} $$ $$ \text{(i)}\ \frac{z_2}{z_1} $$ $$ \text{(j)} \ \overline{ \left ( \frac{z_2}{z_1} \right )} $$ $$\text{(k)} \ \frac{\bar{z}_1z_2}{z_1\bar{z}_2} $$ $$ \text{(l)} \ \frac{z_2}{\bar{z}_1} + \frac{z_1}{\bar{z}_2} $$
2. Let z1 = 3 - 2i, z2 = -1 + 4i. Show that
$$\text{(a)} \ \overline{(z_1 + z_2)} = \bar{z}_1 + \bar{z}_2 $$ $$\text{(b)} \ \overline{z_1 z_2} = \bar{z}_1 \bar{z}_2 $$ $$ \text{(c)} \ \overline{ \left ( \frac{z_1}{z_2} \right )} = \frac{\bar{z}_1}{\bar{z}_2}$$

1.4 Trigonometric Form

As a complex number z = x + yi is an ordered pair (x, y) of real numbers, we can place z in xy-coordinate plane as usual. If the length of the line segment from O to z is r, we say that the absolute value of z is r and is denoted by |z|. Here angle θ, measure in radians, is an angle with positive x-axis and the line segment. From trigonometry, we have known that

x y θ r z = (x, y) = x + yi O

r = √ x2 + y2       x = r cos θ       y = r sin θ


So we have z = (x, y) = x + yi = r cos θ + i r sin θ.
z = r(cos θ + i sin θ)

Which is called the trigonometric form of a complex number. z = (r, θ) is called the polar form of z.

Example 5.
Find the trigonometric form with -π < θ ≤ π.
(a) z = 1 + √ 3 i
(b) z = -1 + i
(c) z = - √ 3 - i
(d) z = -1

Solution

(a) z = 1 + √ 3 i = (1, √ 3 )
r = √ 1 + 3 = 2,       cos θ = 1⁄2       sin θ = √ 3 ⁄ 2
θ = π⁄3
z = 2(cos π⁄3 + i sin π⁄3)
x y θ r z = 1 + √ 3 O
(b) z = -1 + i = (-1, 1)
r = √ 1 + 1 = √ 2 ,       cos θ = - 1 ⁄ √ 2       sin θ = 1⁄ √ 2
θ = 3π⁄4
z = √ 2 (cos 3π⁄4 + i sin 3π⁄4 )
x y θ r z = (-1, 1) O
(c) z = -√ 3 - i = ( -√ 3 - 1)
r = √ 3 + 1 = 2,       cos θ = - √ 3 ⁄2,       sin θ = - 1⁄ 2
θ = - 5π ⁄ 6
z = 2(cos (- 5π ⁄ 6 ) + i sin(- 5π ⁄ 6 ))

x y θ r z = (-√ 3, -1) O

(d) z = -1 = (-1, 0)
r = 1,       cos θ = -1,       sin θ = 0
θ = π
z = cos π + i sin π
x y θ r z = (-1, 0) O

Product in Trigonometric Form
Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2). Then their product is

z1z2 = r1(cos θ1 + i sin θ1) . r2(cos θ2 + i sin θ2)
= r1r2((cos θ1 cos θ2 - sin θ1 sin θ2) + i(sin θ1 cos θ2 + cos θ1 sin θ2))
= r1r2(cos(θ1 + θ2) + i sin((θ1 + θ2))

z1z2 = r1r2(cos(θ1 + θ2) + i sin((θ1 + θ2))


Example 6.
Given that z1 = 1 + √ 3 i, z2 = -1 + i, find z1z2 by using trigonometric forms. Check your answer by direct multiplication.

Solution

z1 = 1 + √ 3 i = 2(cos π ⁄ 3 + i sin π ⁄ 3)                (See Example 5(a))
z2 = -1 + i = √ 2 (cos 3π ⁄ 4 + i sin 3π ⁄ 4)                (See Example 5(b))
z1z2 = 2√ 2   [ cos ( π ⁄ 3 + 3π ⁄ 4) + i sin ( π ⁄ 3 + 3π ⁄ 4)]
= -1 - √ 3 + (1 - √ 3 )i

cos(θ1 + θ2) + i sin((θ1 + θ2) = (cos θ1 cos θ2   +   sin θ1 sin θ2)   +   i(sin θ1 cos θ2 + cos θ1 sin θ2)

2 √ 2 [ (cos π⁄ 3 . cos 3π⁄ 4 - sin π⁄ 3 . sin 3π⁄ 4)   +   i ( sin π⁄ 3 . cos 3π⁄ 4 + cos π⁄ 3 . sin 3π⁄ 4) ]
= 2 √ 2 [ { 1⁄ 2 . (- 1⁄ √ 2 ) - ( √ 3 ⁄2 . 1⁄ √ 2 ) }   +   i { √ 3 ⁄ 2 . (- 1⁄ √ 2 ) + 1⁄ 2 . 1⁄ √ 2 } ]
= 2 √ 2 [ ( - 1⁄ 2 √ 2 - √ 3 ⁄ 2 √ 2 )   +   i ( - √ 3 ⁄2 √ 2 + 1⁄2 √ 2 ) ]
= 2 √ 2 [ ( -1 - √ 3 ⁄ 2 √ 2 )   +   ( - √ 3 + 1⁄ 2 √ 2 ) i ]
= 2 √ 2 ( -1 - √ 3 ⁄ 2 √ 2 )   +   2 √ 2 ( - √ 3 + 1 ⁄ 2 √ 2 ) i
= -1 - √ 3   +   (1 - √ 3 )i

checking
z1z2 = (1 + √ 3 i)(-1 + i)
= -1 + i - √ 3 i + √ 3 i2
= -1 - √ 3 + (1 - √ 3 )i

Multiplicative Inverse in Trigonometric Form
Since z-1 = x ⁄ x2 + y2 - y ⁄ x2 + y2 i for z = x + yi,

z-1 = 1 ⁄ x2 + y2 (x - yi)
= 1 ⁄ r2 r (cos θ - i sin θ)
= 1 ⁄ r(cos θ - i sin θ)

z-1 = 1 ⁄ r(cos (-θ) + i sin (-θ))


Example 7.
Given that z = -√ 3 - i, using trigonometric form of z, find z-1. Check your answer by showing that zz-1 = 1.

Solution

z = -√ 3 - i
z = 2(cos (-5π ⁄ 6) + i sin (- 5π ⁄ 6))                   (See Example 5(c))
z-1 = 1 ⁄ 2(cos 5π ⁄ 6 + i sin (5π ⁄ 6))
= 1 ⁄ 2 ( - √ 3 ⁄ 2 + 1 ⁄ 2 i)
= - √ 3 ⁄ 4 + 1 ⁄ 4 i
zz-1 = (-√ 3 - i) (- √ 3 ⁄ 4 + 1 ⁄ 4 i) = 3 ⁄ 4 - √ 3 ⁄ 4 i + √ 3 ⁄ 4 i - 1 ⁄ 4 i2 = 3 ⁄ 4 + 1 ⁄ 4 = 1

Division in Trigonometric Form
Let z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2). Then

z1 ⁄ z2 = z1z2-1
= r1(cos θ1 + i sin θ1) 1 ⁄ r2(cos(-θ2) + i sin(- θ2) + i sin(-θ2))
z1 ⁄ z2 =r1 ⁄ r2 (cos(θ1 - θ2) + i sin(θ1 - θ2))


Example 8.

Given that z1 = 1 +√ 3 i, z2 = -1 + i, find z1 ⁄ z2 by using trigonometric form. Check your answer by direct solution.

Solution

z1 = 1 +√ 3 i = 2(cos π ⁄ 3 + i sin π ⁄ 3)
z2 = -1 + i = √ 2 (cos 3π ⁄ 4 + i sin 3π ⁄ 4)
z1 ⁄ z2 = 2 ⁄ √ 2 (cos ( π ⁄ 3 - 3π ⁄ 4) + i sin ( π ⁄ 3 - 3π ⁄ 4))
= -1 + √ 3 ⁄ 2 - 1 + √ 3 ⁄ 2 i
z1 ⁄ z2 = 1 + √ 3 i ⁄ -1 + i -1 - i⁄ -1 - i
= -1 - i - √ 3 i - √ 3 i2 ⁄ 1 + 1
= -1 + √ 3 - (1 + √ 3 ) i ⁄ 2
= -1 + √ 3 ⁄ 2 - 1 + √ 3 ⁄ 2 i

Powers of Complex Numbers
The power of a complex number z = r(cos θ + i sin θ) is given by
zn = rn (cos nθ +i sin nθ),     n in an integer


Example 9.
Given that z = 1 + √ 3 i, find (a) z10       (b) z-10

Solution

z = 1 + √ 3 i = 2(cos π ⁄ 3 + i sin π ⁄ 3)               (See Example 5(a))
(a) z10 = 210 (cos 10 π ⁄ 3 + i sin 10 π ⁄ 3) = 1024 (- 1 ⁄ 2 - √ 3 ⁄ 2 i) = -512 - 512 √ 3 i

(b) z-10 = 1 ⁄ 210 (cos (- 10 π ⁄ 3) + i sin (- 10 π ⁄ 3)) = 1 ⁄ 1024 (- 1 ⁄ 2 + √ 3 ⁄ 2 i) = - 1 ⁄ 2048 + √ 3 ⁄ 2048 i

Exercise 1.4

  1. Find the trigonometric form with -π < θ ≤ π.
    2. (a) z = 1 - √ 3 i
    (b) z = - √ 2 + √ 2 i
    (c) z = -2 - 2i
    (d) z = √ 3 - i
    (e) z = i
    (f) z = -3i
  2. Given that z1 = 2 - 2√ 3 i, z2 = -1 - i, find the following complex numbers by using trigonometric forms. Check your answer by direct calculation.
    3. (a) z1z2
    (b) z1-1
    (c) z2-1
    (d) z1 ⁄ z2
    (e) z2 ⁄ z1
  3. Given that z = -2 √ 3 - 2i, find (a) z5       (b) z-5

1.5 Roots of Complex Numbers

An nth root of a complex number z is a complex number ω that satisfies the equation
ωn = z

Since z = r(cos θ + i sin θ) and
$$ (\sqrt[n]{r}\ (\text{cos}\ \frac{\theta}{n} + i \ \text{sin}\ \frac{\theta}{n}))^n $$ $$= (\sqrt[n]{r})^n (\text{cos}\ n (\frac{\theta}{n}) + i\ \text{sin}\ n (\frac{\theta}{n})) $$ $$ = r (\text{cos}\ \theta + i\ \text{sin}\ \theta) = z $$ We have an nth root of a complex number z as
$$ \sqrt[n]{r} (\text{cos}\ \frac{\theta}{n} + i\ \text{sin}\ \frac{\theta}{n}). $$ Moreover, since z can also be represented as

z = r(cos (θ + 2kπ) + i sin ( θ + 2kπ))

for every integer k, we have
$$\small{ (\sqrt[n]{r} (\text{cos}\ \frac{\theta + 2k\pi}{n} + i\ \text{sin}\ \frac{\theta + 2k\pi}{n}))^n} $$ $$\small{= (\sqrt[n]{r})^n (\text{cos}\ n\ \frac{\theta + 2k\pi}{n} + i\ \text{sin}\ n\ \frac{\theta + 2k\pi}{n}) }$$ $$ = r (\text{cos}\ (\theta + 2k\pi) + i\ \text{sin}\ (\theta + 2k\pi )) $$ $$ = z$$ So nth roots of z are
$$ \sqrt[n]{r} (\text{cos}\ \frac{\theta + 2k\pi}{n} + i\ sin\ \frac{\theta + 2k\pi}{n}) $$ for every integer k

But for k = n, we have
cos θ + 2kπ ⁄ n + i sin θ + 2kπ ⁄ n = cos θ + 2nπ ⁄ n + i sin θ + 2nπ ⁄ n
=cos (θ ⁄ n + 2π) + i sin( θ ⁄ n + 2π)
= cos θ ⁄ n + i sin θ ⁄ n
so that root for k = n is the same as the root for k = 0. The same is also true for k > n as the root for k = n + 1 is same as the root for k = 1, and so on. Therefore the roots of z, denoted by ωk of z are
$$ \omega_k = \sqrt[n]{r} (\text{cos}\ \frac{\theta + 2k\pi}{n} + i \ \text{sin}\ \frac{\theta + 2k\pi}{n}), \ \ \ \ \ k = 1,2,..., n-1. $$ Example 10.
Find the cube roots of z = -2 - 2i.

Solution

z = -2 - 2i = (-2, -2)
r = √ 4 + 4 = 2 √ 2 ,     cos θ = - 1 ⁄ √ 2 ,     sin θ = - 1 ⁄ √ 2
θ = - 3π ⁄ 4
z = 2 √ 2 (cos (- 3π ⁄ 4) + i sin ( - 3π ⁄ 4))

x y θ r z = (-2, -2) O
Then we have
ωk = ∛ 2 √ 2 (cos - 3π ⁄ 4 + 2kπ ⁄ 3 + i sin - 3π ⁄ 4 + 2kπ ⁄ 3),       k = 0, 1, 2.
Therefore the cube roots are
ω0 = √ 2 (cos(- 3π ⁄ 12) + i sin (- 3π ⁄ 12))
= √ 2 ( √ 2 ⁄ 2 - i √ 2 ⁄ 2)
= 1 - i
ω1 = √ 2 (cos 5π ⁄ 12 + i sin 5π ⁄ 12)
= √ 2 ( √ 6 - √ 2 ⁄ 4 + i √ 6 + √ 2 ⁄ 4)
= -1 + √ 3 ⁄ 2 + 1 + √ 3 ⁄2 i
ω2 = √ 2 (cos( 13π ⁄ 12) + i sin ( 13π ⁄ 12))
= √ 2 (- √ 6 + √ 2 ⁄ 4 + i - √ 6 + √ 2 ⁄ 4)
= - 1 + √ 3 ⁄ 2 + 1 - √ 3 ⁄2 i

Example 11.

Solve z6 = 1.

Solution

1 = (1, 0) = 1(cos 0 + i sin 0) = 1 (cos (0 + 2kπ) + i sin(0 + 2kπ))
z6 = 1
26 = 1 (cos (0 + 2kπ) + i sin(0 + 2kπ))
z =1 (cos (0 + 2kπ) + i sin(0 + 2kπ))1 ⁄ 6
z = 6√ 1 (cos 0 + 2kπ ⁄ 6 + i sin 0 + 2kπ ⁄ 6),         k = 0, 1, 2, ....., 5

When k = 0, z = cos 0 + i sin 0 = 1.

When k = 1, z = cos π ⁄ 3 + i sin π ⁄ 3 = 1 ⁄ 2 + i √ 3 ⁄ 2 .

When k = 2, z = cos 2π ⁄ 3 + i sin 2π ⁄ 3 = - 1 ⁄ 2 + i √ 3 ⁄ 2 .

When k = 3, z = cos π + i sin π = -1.

When k = 4, z = cos 4π ⁄ 3 + i sin 4π ⁄ 3 = - 1 ⁄ 2 - i √ 3 ⁄ 2 .

When k = 5, z = cos 5π ⁄ 3 + i sin 5π ⁄ 3 = 1 ⁄ 2 - i √ 3 ⁄ 2 .


Exercise 1.5

  1. Find the square roots of the following complex numbers.

    2. (a) 1 + √ 3 i
    (b) i
    (c) - √ 3 + i
    (d) -1 - √ 3 i
    (e) - i
    (f) √ 3 - i

  2. Find the cube roots of the following complex numbers.

    3. (a) 1 + i
    (b) i
    (c) -1 + i
    (d) -1 - i
    (e) -i
    (f) 1 - i

  3. Solve the following equations.

    4. (a) z4 = -i
    (b) z4 = -1
    (c) z4 = -8 - 8 √ 3 i
    (d) z6 = -1


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