Complex Numbers

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 + y²}i

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:

(a) z₁² - 2z₁ + 1
(b) 3z₂² + 2z₂ - 1
$$ \text{(c)}\ z_1\bar{z}_2 + z_2\bar{z}_1 $$ (d) ¹⁄_z₁
(e) ¹⁄_z₂
(f) ¹⁄_z₁z₂
(g) ^z₁⁄_z₂
$$ \text{(h)}\ \frac{\bar{z}_1}{\bar{z}_2} $$ (i) ^z₂⁄_z₁
$$ \text{(j)} \ \overline{ \left ( \frac{z_2}{z_1} \right )} $$ $$\text{(k)} \ \frac{\bar{z}_1z_2}{z_1\bar{z}_2} $$ $$ \text{(l)} \ \frac{z_2}{\bar{z}_1} + \frac{z_1}{\bar{z}_2} $$ ပံ့ပိုးကူညီသူများ

Answers
1. z₁ = -2 +3i, z₂ = 5 + 2i

(a) z₁² - 2z₁ + 1
= (-2 + 3i)² - 2 (-2 + 3i) + 1
= 4 - 12i + 9i² + 4 - 6i + 1
= 9 - 18i + 9(-1)
= 9 - 18i - 9
= -18i

(b) 3z₂² + 2z₂ - 1
= 3 (5 + 2i)² + 2 (5 + 2i) - 1
= 3 (25 + 20i + 4i²) + 10 + 4i - 1
= 75 + 60i + 12i² + 10 + 4i - 1
= 84 + 64i + 12(-1)
= 84 + 64i - 12
= 72 + 64i

(c) z₁ z₂ + z₂ z₁ = (-2 + 3i) (5 - 2i) + (5 + 2i) (-2 - 3i)
= (-10 + 4i + 15i - 6i²) + (-10 - 15i - 4i - 6i²)
= (-10 + 19i - 6(-1) ) + (-10 - 19i - 6(-1) )
= (-4 + 19i) + (-4 - 19i)
= -8 + 0i
= -8

(d) ¹⁄ _z₁
= ¹⁄ _{-2 + 3i} × ^{-2 - 3i}⁄ _{-2 - 3i}
= ^{-2 - 3i}⁄ _{4 - 9i²}
= ^{-2 - 3i} ⁄ _{4 - 9(-1)}
= ^{-2 - 3i} ⁄ _{4 + 9}
= ^{-2 - 3i}⁄ ₁₃
= ^-2⁄ ₁₃ - ³⁄ ₁₃ i

(e) ¹⁄ _z₂ = ¹⁄ _{5 + 2i} × ^{5 - 2i}⁄ _{5 - 2i}
= ^{5 - 2i} ⁄ _{25 - 10i + 10i - 4i²}
= ^{5 - 2i} ⁄ _{25 - 4i²}
= ^{5 - 2i} ⁄ _{25 - 4(-1)}
= ^{5 - 2i}⁄ _{25 + 4}
= ^{5 - 2i}⁄ ₂₉
= ⁵⁄ ₂₉ - ²⁄ ₂₉ i

(f) ¹⁄ _z₁z₂ = ¹⁄ _{(-2 + 3i) (5 + 2i)}

= ¹⁄ _{-10 - 4i + 15i + 6i²}
= ¹⁄ _{-10 + 11i + 6(-1)}
= ¹⁄_{-16 + 11i}
= ¹⁄ _{-16 + 11i} × ^{-16 - 11i}⁄ _{-16 - 11i}
= ^{-16 - 11i}⁄_{256 + 176i - 176i - 121i²}
= ^{-16 - 11i}⁄ _{256 - 121(-1)}
= ^{-16 - 11i}⁄ _{256 + 121}
= ^{-16 - 11i}⁄ ₃₇₇
= ^-16⁄ ₃₇₇ - ¹¹⁄ ₃₇₇ i

(g) ^z₁⁄ _z₂ = ^{-2 + 3i}⁄ _{5 + 2i} × ^{5 - 2i}⁄_{5 - 2i}
= ^{-10 + 4i + 15i - 6i²}⁄ _{25 - 10i + 10i - 4i²}
= ^{-10 + 19i - 6(-1)}⁄_{25 - 4(-1)}
= ^{-10 + 19i + 6}⁄_{25 + 4}
= ^{-4 + 19i}⁄ ₂₉
= ^-4⁄ ₂₉ + ¹⁹⁄ ₂₉ i

(h) ^z₁⁄ _z₂ = ^{-2 - 3i}⁄ _{5 - 2i} × ^{5 + 2i}⁄_{5 + 2i}
= ^{-10 - 4i - 15i - 6i²}⁄ _{25 + 10i - 10i - 4i²}
= ^{-10 - 19i - 6(-1)}⁄_{25 - 4(-1)}
= ^{-10 - 19i + 6}⁄_{25 + 4}
= ^{- 4 - 19i}⁄ ₂₉
= ^{- 4}⁄ ₂₉ - ¹⁹⁄ ₂₉ i

(i) ^z₂⁄ _z₁ = ^{5 + 2i}⁄ _{-2 + 3i} × ^{-2 - 3i}⁄_{-2 - 3i}
= ^{-10 - 15i - 4i - 6i²}⁄_{4 + 6i - 6i - 9i²}
= ^{-10 - 19i - 6(-1)}⁄ _{4 - 9(-1)}
= ^{-10 - 19i + 6}⁄_{4 + 9}
= ^{- 4 - 19i}⁄ ₁₃
= ^{- 4}⁄ ₁₃ - ¹⁹⁄ ₁₃ i

$$\text{(j)}\ \ \overline{(\frac{z_2}{z_1})} = \overline{\left(\frac{5+2i}{-2+3i} \times \frac{-2-3i}{-2-3i}\right)} $$ $$ = \overline{\left(\frac{-10 - 15i - 4i - 6i^2}{4 + 6i -6i - 9i^2}\right)} $$ $$ = \overline{\left(\frac{-10 - 19i - 6(-1)}{4 - 9(-1)}\right)} $$ $$ = \overline{\left(\frac{-4 - 19i}{13}\right)} $$ $$ = \frac{-4-19i}{13} $$ $$ = \frac{-4}{13} - \frac{19}{13}i $$
(k) ^{z₁ z₂}⁄_{z₁ z₂}
= ^{(-2 - 3i) (5 + 2i)}⁄ _{(-2 + 3i) (5 - 2i)}
= ^{-10 - 4i - 15i - 6i²}⁄ _{-10 + 4i + 15i - 6i²}
= ^{-10 - 19i - 6(-1)}⁄_{-10 + 19i - 6(-1)}
= ^{-10 - 19i + 6}⁄_{-10 + 19i + 6}
= ^{-4 - 19i}⁄ _{-4 + 19i}
= ^{-4 - 19i}⁄ _{-4 + 19i} × ^{-4 - 19i}⁄ _{-4 - 19i}
= ^{16 + 76i + 76i + 361i²}⁄ _{16 + 76i - 76i - 361i²}
= ^{16 + 152i + 361(-1)}⁄ _{16 - 361(-1)}
= ^{16 + 152i - 361}⁄ _{16 + 361}
= ^{-345 + 152i}⁄ ₃₇₇
= ^-345⁄ ₃₇₇ + ¹⁵²⁄ ₃₇₇ i

(l) ^z₂⁄ _z₁ + ^z₁⁄ _z₂ = ^{z₂ z₂ + z₁ z₁}⁄ _{z₁ z₂}
= ^{(5 + 2i) (5 - 2i) + (-2 + 3i (-2 - 3i)}⁄ _{(-2 - 3i) (5 - 2i)}
= ^{(25 - 10i + 10i - 4i²) + (4 + 6i - 6i - 9i²)}⁄_{-10 + 4i - 15i + 6i²}
= ^{(25 - 4(-1) ) + (4 - 9(-1) )}⁄_{-10 - 11i + 6(-1)}
= ^{25 + 4 + 4 + 9}⁄ _{-10 - 11i - 6}
= ⁴²⁄_{-16 - 11i}
= ⁴²⁄ _{-16 - 11i} × ^{-16 + 11i}⁄ _{-16 + 11i}
= ^{-672 + 642i}⁄ _{256 + 121}
= ^{-672 + 462i}⁄ ₃₇₇
= ^-672⁄ ₃₇₇ + ⁴⁶²⁄ ₃₇₇ i

2. z₁ = 3 - 2i, z₂ = -1 + 4i
(a) (z₁ + z₂) = z₁ + z₂
(z₁ + z₂) = ( (3 - 2i) + (-1 + 4i) )
= (2 + 2i)
= (2 - 2i)

z₁ + z₂ = (3 - 2i) + (-1 + 4i) = (3 + 2i) + (-1 - 4i)

∴ (z₁ + z₂) = z₁ + z₂

(b) z₁ z₂ = z₁ z₂

z₁ z₂ = (3 - 2i) (-1 + 4i)
= (-3 + 12i + 2i - 8i²)
= (-3 + 14z - 8(-1) )
(-3 + 14i + 8)
= (5 + 14z)
= (5 - 14i)

z₁ z₂ = (3 - 2i) (-1 + 4i)
= (3 + 2i) (-1 - 4i)
= (-3 -12i - 2i - 8i²
= -3 - 14i - 8(-1)
= -3 - 14i + 8
= 5 - 14i
∴ z₁ z₂ = z₁ z₂

(c)
$$ \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}} $$ $$ \overline{\left(\frac{z_1}{z_2}\right)} = \overline{\left(\frac{3 - 2i}{-1 + 4i}\right) \times \left(\frac{-1 - 4i}{-1 - 4i}\right)} $$ $$ = \overline{\left(\frac{-3 - 12i + 2i + 8i^2}{1 + 4i - 4i - 16i^2}\right)} $$ $$ = \overline{\left(\frac{-3 - 10i + 8(-1)}{1 - 16(-1)}\right)} $$ $$ = \overline{\left(\frac{-11 - 10i}{17}\right)} $$ $$ = \frac{-11 + 10i}{17} $$
$$ \frac{\overline{z_1}}{\overline{z_2}} = \frac{\overline{(3 - 2i)}}{\overline{(-1 + 4i)}} $$ $$ = \frac{(3 + 2i)}{(-1 - 4i)} $$ $$ = \frac{3 + 2i}{-1 - 4i} \times \frac{-1 + 4i}{-1 + 4i} $$ $$ = \frac{-3 + 12i - 2i + 8i^2}{1 - 4i + 4i - 16i^2} $$ $$ = \frac{-3 + 10i + 8(-1)}{1 - 16(-1)} $$ $$ = \frac{-11 + 10i}{17} $$ $$ \therefore \overline{\left(\frac{z_1}{z_2}\right)} = \frac{\overline{z_1}}{\overline{z_2}} $$

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