Operations with Complex Numbers

Complex numbers follow the same fundamental operations as real numbers: addition, multiplication, subtraction, and division.

Adding Complex Numbers

Given two complex numbers \((x+iy)\) and \((x'+iy')\), their sum is calculated as: $$ (x+iy) + (x'+iy') = (x+x')+i(y+y') $$

The process is straightforward: simply add the real parts together and the imaginary parts together, just as you would with real numbers.

Example

Consider these two complex numbers:

$$ 3+i2 $$ $$ 4+i7 $$

Adding them step by step:

$$ (3+i2) + (4+i7) = $$

$$ 3+i2 + 4+i7 = $$

$$ (3+4)+i(2+7) = $$

$$ = 7+i9 $$

Note: A quicker way to add complex numbers is by treating them as coordinate pairs in the complex plane and adding the corresponding components. $$ (3+i2) + (4+i7) = $$ $$ (3,2)+(4,7) $$ $$ (3+4,2+7) $$ $$ (7,9) $$ These coordinates correspond to the complex number $$ 7+i9 $$ The result is the same.

Subtracting Complex Numbers

Given two complex numbers \((x+iy)\) and \((x'+iy')\), their difference is calculated as: $$ (x+iy) - (x'+iy') = (x-x')+i(y-y') $$

The approach is similar to addition—subtract the real parts and the imaginary parts separately, following the usual algebraic rules.

Example

Consider these two complex numbers:

$$ 4+i7 $$ $$ 3+i2 $$

Subtracting them step by step:

$$ (4+i7) - (3+i2) = $$

$$ 4+i7 -3-i2 = $$

$$ (4-3)+i(7-2) = $$

$$ = 1+i5 $$

Note: Just like with addition, we can represent the complex numbers as coordinate pairs and subtract their components separately. $$ (4+i7) - (3+i2) = $$ $$ (4,7)-(3,2) $$ $$ (4-3,7-2) $$ $$ (1,5) $$ These coordinates correspond to the complex number $$ 1+i5 $$ The result remains the same.

Multiplying Complex Numbers

The product of two complex numbers follows this formula: $$ (x+iy) \cdot (x'+iy') = (xx'-yy')+i(xy'+x'y) $$

Unlike addition and subtraction, multiplication requires an understanding of the imaginary unit properties: \( i = \sqrt{-1} \) and \( i^2 = -1 \).

Note: The multiplication formula changes when working with trigonometric form: $$ z \cdot z' = (d \cdot d') \cdot [ \cos(a+a') + i \cdot \sin(a+a') ] $$ In exponential form, the multiplication is expressed as: $$ z \cdot z' = ( d \cdot e^{ia } ) \cdot ( d' \cdot e^{ia'} ) = ( d \cdot d' ) \cdot e^{i(a+a')} $$

Example

Consider these two complex numbers:

$$ 4+i7 $$ $$ 3+i2 $$

Multiplying them step by step:

$$ (4+i7) \cdot (3+i2) = $$

Using the multiplication formula:

$$ (x+iy) \cdot (x'+iy') = (xx'-yy')+i(xy'+x'y) = $$

$$ (4+i7) \cdot (3+i2) = ((4 \cdot 3)-(7 \cdot 2))+i((4 \cdot 2)+(3 \cdot 7)) = $$

$$ (12-14)+i(8+21) = $$

$$ = -2+i29 $$

Note: The same result can be obtained by expanding the expression using standard algebraic rules and substituting \( i^2 = -1 \), without relying directly on the multiplication formula. $$ (4+i7) \cdot (3+i2) = $$ $$ (4 \cdot 3 ) + (4 \cdot i2) + (i7 \cdot 3 ) + (i7 \cdot i2) $$ $$ 12 + i8 + i21 + i^2 14 = $$ $$ 12 + i8 + i21 + (-1) \cdot 14 = $$ $$ (12-14) + i (8+21) = $$ $$ = -2 + i \cdot 29 $$ The result is the same.

Dividing Complex Numbers

Given two complex numbers \( x+iy \) and \( x'+iy' \), as long as the denominator is nonzero, division is performed using the following formula: $$ \frac{x+iy}{x'+iy'} = \frac{xx'+yy'}{x'^2+y'^2}+i \cdot \frac{x'y-xy'}{x'^2+y'^2} $$ Alternatively, division can be carried out by multiplying both the numerator and denominator by the conjugate of the denominator: $$ \frac{x+iy}{x'+iy'} = \frac{x+iy}{x'+iy'} \cdot \frac{x'-iy'}{x'-iy'} $$

In trigonometric form, division follows this formula:

$$ \frac{z}{z'} = \frac{d}{d'} \cdot [ \cos (a-a') + i \cdot \sin (a-a') ] $$

For the exponential form, it is expressed as:

$$ \frac{z}{z'} = \frac{d \cdot e^{ia }}{d' \cdot e^{ia'}} = \left( \frac{d}{d'} \right) \cdot e^{i(a-a')} $$

Example

Let’s divide the following two complex numbers:

$$ 3+i2 $$ $$ 4+i7 $$

Using the division formula:

$$ \frac{ xx'+yy' }{ x'^2 + y'^2 } + i \cdot \frac{x'y-xy'}{x'^2 + y'^2} = $$

$$ \frac{ (3 \cdot 4)+ (2 \cdot 7) }{ 4^2 + 7^2 } + i \cdot \frac{(4 \cdot 2) - (3 \cdot 7) }{4^2 + 7^2} = $$

$$ \frac{ 12+14 }{ 16 + 49 } + i \cdot \frac{8 - 21 }{16 + 49} = $$

$$ = \frac{26 }{ 65 } + i \cdot \frac{-13 }{65} $$

$$ = \frac{26 }{ 65 } - i \cdot \frac{13 }{65} $$

Note: The same result can be obtained by using algebraic properties and the imaginary unit \( i^2=-1 \). $$ \frac{3+i2}{4+i7} = $$ Multiply both the numerator and denominator by the conjugate of the denominator: $$ \frac{3+i2}{4+i7} \cdot \frac{4-i7}{4-i7} = $$ Expanding: $$ \frac{(3+i2) \cdot (4-i7) }{(4+i7) \cdot (4-i7) } = $$ $$ \frac{12-i21+i8-i^214 }{ 16-i28+i28-i^249 } = $$ Since \( i^2=-1 \), we simplify: $$ \frac{12-i21+i8-(-1) \cdot 14 }{ 16-i28+i28-(-1) \cdot 49 } = $$ $$ \frac{12-i21+i8 +14 }{ 16+49} = $$ $$ \frac{(12+14)+i(8-21) }{ 16+49} = $$ $$ \frac{26+i(-13) }{65} = $$ $$ = \frac{26 }{ 65 } - i \cdot \frac{13 }{65} $$

Raising a Complex Number to a Power

The power of a complex number \( z=x+yi \) can be computed using binomial expansion, keeping in mind that \( i^2=-1 \): $$ z^n = (x+yi)^2 $$

In trigonometric form, exponentiation follows the rule:

$$ z^n = d^n \cdot ( \cos (n \cdot a) + i \cdot \sin (n \cdot a) ) $$

In exponential form, it is expressed as:

$$ z^n = d^n \cdot e^{ia \cdot n } $$

Example

Let's calculate the square of the complex number \( z=1+3i \):

$$ z^2 = (1+3i)^2 $$

Expanding using the binomial formula:

$$ z^2 = 1^2+2 \cdot 3i + (3i)^2 $$

$$ z^2 = 1+6i + 9i^2 $$

Since \( i^2=-1 \), we substitute:

$$ z^2 = 1+6i + 9 \cdot (-1) $$

$$ z^2 = 1+6i -9 $$

$$ z^2 = -8+6i $$

Thus, the square of the complex number is: \( z^2=(1+3i)^2=-8+6i \).

Note: The exponentiation of a complex number can also be calculated using its trigonometric form: $$ z = \sqrt{10} \cdot ( \cos 71.57^\circ + i \cdot \sin 71.57^\circ ) $$ Applying the exponentiation formula with \( n=2 \): $$ z^n = d^n \cdot ( \cos (n \cdot a) + i \cdot \sin (n \cdot a) ) $$ $$ z^2 = ( \sqrt{10} )^2 \cdot ( \cos (2 \cdot 71.57^\circ) + i \cdot \sin (2 \cdot 71.57^\circ) ) $$ $$ z^2 = 10 \cdot ( \cos (143.14^\circ) + i \cdot \sin (143.14^\circ) ) $$ $$ z^2 = -8 + i 6 $$

Square of a Complex Number

The square of a complex number \((x,y)^2\) can also be computed using the following formula: $$ (x,y)^2 = (x^2-y^2,2xy) $$

Example

Let's find the square of the complex number:

$$ z = (2,3) $$

Using the formula:

$$ z^2 = (x,y)^2 = (x^2-y^2,2xy) $$

Substituting \( x=2 \) and \( y=3 \):

$$ (2,3)^2 = (2^2-3^2,2 \cdot 2 \cdot 3) $$

$$ (2,3)^2 = (4-9,12) $$

$$ (2,3)^2 = (-5,12) $$

Thus, the square of the complex number is \( (-5,12) \).

Expressed in algebraic form:

$$ z^2 = -5 + 12 i $$

Verification: Let's confirm the result by computing the square in algebraic form. Given that \( z=(2,3)=2+3i \): $$ z^2 = (2+3i)^2 $$ $$ z^2 = 2^2+2 \cdot 2 \cdot 3i + (3i)^2 $$ $$ z^2 = 4+12i + 9i^2 $$ Since \( i^2=-1 \), we substitute: $$ z^2 = 4+12i + 9 \cdot (-1) $$ $$ z^2 = 4+12i -9 $$ $$ z^2 = -5+12i $$ The result is the same.

Cube of a Complex Number

To compute the cube of a complex number in algebraic form, we use the binomial expansion formula: $$ z^3=(a+bi)^3=a^3+3a^2(bi)+3a(bi)^2+(bi)^3 $$ remembering that \( i^2=-1 \).

Example

Let's compute the cube of the complex number \( z=2+3i \):

$$ z^3 = (2+3i)^3 $$

Expanding the binomial:

$$ z^3 = 2^3+3 \cdot 2^2 \cdot 3i + 3 \cdot 2 \cdot (3i)^2 + (3i)^3 $$

$$ z^3 = 8+36i + 54 i^2 + 27i^3 $$

Since \( i^2=-1 \):

$$ z^3 = 8+36i + 54 \cdot (-1) + 27i^3 $$

$$ z^3 = 8+36i - 54 + 27i^3 $$

$$ z^3 = -46+36i + 27i^3 $$

Knowing that \( i^3=i^2 \cdot i \) and \( i^2=-1 \):

$$ z^3 = -46+36i + 27 \cdot (-1) \cdot i $$

$$ z^3 = -46+36i - 27 i $$

$$ z^3 = -46+9i $$

Thus, the cube of the complex number is \( z^3=-46+9i \).

Root of a Complex Number

To compute the \( n \)th root of a complex number, we first express it in trigonometric form: $$ z = d \cdot [ \cos a + i \cdot \sin a ] $$ and then apply the following formula: $$ \sqrt[n]{z} = \sqrt[n]{d} \cdot \left[ \cos \frac{(a+2\pi k)}{n} + i \cdot \sin \frac{(a+2\pi k) }{n} \right] $$ for \( k \) ranging from 0 to \( n-1 \).

Using trigonometric form, the \( n \)th root is calculated as:

$$ \sqrt[n]{z} = \sqrt[n]{d} \cdot \left[ \cos \frac{(a+2\pi k)}{n} + i \cdot \sin \frac{(a+2\pi k) }{n} \right] $$

In exponential form, it is expressed as:

$$ \sqrt[n]{z} = d^{\frac{1}{n}} \cdot e^{i \cdot \frac{(a+2\pi k)}{n}} $$

Example

Consider the complex number:

$$ z = -8 + i 6 $$

Expressed in trigonometric form:

$$ z = 10 \cdot ( \cos (143.14^\circ) + i \cdot \sin (143.14^\circ) ) $$

In radians:

$$ z = 10 \cdot ( \cos (2.49 \text{ rad}) + i \cdot \sin (2.49 \text{ rad}) ) $$

Computing the square root (\( n=2 \)), which has two solutions:

For \( k=0 \):

$$ \sqrt[2]{z} = \sqrt[2]{10} \cdot \left[ \cos \frac{2.49}{2} + i \cdot \sin \frac{ 2.49 }{2} \right] $$

$$ = 1 + 3 i $$

For \( k=1 \):

$$ \sqrt[2]{z} = \sqrt[2]{10} \cdot \left[ \cos \frac{ (2.49+2\pi)}{2} + i \cdot \sin \frac{ (2.49+2\pi) }{2} \right] $$

$$ = - 1 - 3 i $$

Reciprocal of a Complex Number

The reciprocal of a complex number \( z_1 \) is another complex number \( z_2 \) such that their product equals 1: $$ z_1 \cdot z_2 = 1 $$

If \( z_1 \neq 0 \), its reciprocal is given by:

$$ z_2 = \frac{1}{z_1} = \frac{\overline{z_1}}{|z_1|^2} $$

where \( \overline{z_1} \) is the complex conjugate of \( z_1 \) and \( |z_1|^2 \) is its squared modulus: \( z_1 \cdot \overline{z_1} \).

For example, if \( z_1 = a + bi \), where \( a, b \in \mathbb{R} \), its reciprocal is: $$ z_2 = \frac{a - bi}{a^2 + b^2} $$