1.4 Complex Numbers in Different Forms | Complex Numbers | Mathematics 11

Polar Coordinate System

The polar coordinate system represents a point $P$ in the plane using an ordered pair $(r, θ)$ , where:

$r$ = radial distance from the pole (origin)
$θ$ = angle measured counterclockwise from the polar axis (positive x-axis)

Conversion between Polar and Cartesian coordinates: $x = r cos θ, y = r sin θ$ $r = x^{2} + y^{2}, θ = tan^{- 1} (\frac{y}{x}) (adjusted for quadrant)$

Polar Form of a Complex Number

A complex number $z = x + i y$ can be written in polar form as: $z = r (cos θ + i sin θ)$

where:

$r = ∣ z ∣ = x^{2} + y^{2}$ is the modulus
$θ = ar g (z)$ is the argument

Using Euler's formula: $e^{i θ} = cos θ + i sin θ$ , the polar form becomes: $z = r e^{i θ}$

Quadrant Adjustment for Argument

Quadrant	Condition	$θ$
I	$x > 0, y > 0$	$tan^{- 1} (\frac{y}{x})$
II	$x < 0, y > 0$	$\pi - \tan^{-1}!\left(\frac{
III	$x < 0, y < 0$	$-\pi + \tan^{-1}!\left(\frac{
IV	$x > 0, y < 0$	$-\tan^{-1}!\left(\frac{

Example: Convert $z = - 2 - 2 i$ to polar form.

$r = (- 2)^{2} + (- 2)^{2} = 8 = 22$
Reference angle: $α = tan^{- 1} (1) = \frac{π}{4}$
Since $z$ is in Quadrant III: $θ = - π + \frac{π}{4} = - \frac{3 π}{4}$
Polar form: $z = 22 (cos (- \frac{3 π}{4}) + i sin (- \frac{3 π}{4}))$

Operations with Complex Numbers in Polar Form

If $z_{1} = r_{1} e^{i θ_{1}}$ and $z_{2} = r_{2} e^{i θ_{2}}$ : $z_{1} z_{2} = r_{1} r_{2} e^{i (θ_{1} + θ_{2})}$ $∣ z_{1} z_{2} ∣ = ∣ z_{1} ∣∣ z_{2} ∣, ar g (z_{1} z_{2}) = ar g (z_{1}) + ar g (z_{2})$

Division

$\frac{z _{1}}{z _{2}} = \frac{r _{1}}{r _{2}} e^{i (θ_{1} - θ_{2})}$ $\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}, ar g (\frac{z _{1}}{z _{2}}) = ar g (z_{1}) - ar g (z_{2})$

De Moivre's Theorem

$z^{n} = r^{n} (cos n θ + i sin n θ) = r^{n} e^{in θ}$

Modulus of a Product (Key Identity)

If $(x_{1} + i y_{1}) (x_{2} + i y_{2}) \dots (x_{n} + i y_{n}) = a + ib$ , then: $(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2}) \dots (x_{n}^{2} + y_{n}^{2}) = a^{2} + b^{2}$

Proof: Take modulus of both sides and square: $∣ z_{1} ∣∣ z_{2} ∣ \dots ∣ z_{n} ∣ = ∣ a + ib ∣$ , so $∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2} \dots ∣ z_{n} ∣^{2} = ∣ a + ib ∣^{2}$ .

Equations and Identities in Polar Form

Using $a + b + c = 0$ to Prove Trigonometric Identities

If $cos α + cos β + cos γ = 0$ and $sin α + sin β + sin γ = 0$ , then:

Let $a = e^{i α}$ , $b = e^{i β}$ , $c = e^{iγ}$ . The conditions imply $a + b + c = 0$ .

Using the algebraic identity: $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$

Since $a + b + c = 0$ : $a^{3} + b^{3} + c^{3} = 3 ab c$

Substituting back: $e^{i 3 α} + e^{i 3 β} + e^{i 3 γ} = 3 e^{i (α + β + γ)}$

Taking real parts: $cos 3 α + cos 3 β + cos 3 γ = 3 cos (α + β + γ)$

Real-World Application: Complex Impedance

In AC circuit analysis, complex impedance $Z$ relates voltage $E$ and current $I$ : $Z = \frac{E}{I}$

To simplify division of complex numbers, multiply by the conjugate of the denominator: $Z = \frac{E}{I} = \frac{E \cdot I ˉ}{I \cdot I ˉ} = \frac{E I ˉ}{∣ I ∣ ^{2}}$

Polar Coordinate System

The polar coordinate system represents a point $P$ in the plane using an ordered pair $(r, θ)$ , where:

$r$ = radial distance from the pole (origin)
$θ$ = angle measured counterclockwise from the polar axis (positive x-axis)

Conversion between Polar and Cartesian coordinates: $x = r cos θ, y = r sin θ$ $r = x^{2} + y^{2}, θ = tan^{- 1} (\frac{y}{x}) (adjusted for quadrant)$

Polar Form of a Complex Number

A complex number $z = x + i y$ can be written in polar form as: $z = r (cos θ + i sin θ)$

where:

$r = ∣ z ∣ = x^{2} + y^{2}$ is the modulus
$θ = ar g (z)$ is the argument

Using Euler's formula: $e^{i θ} = cos θ + i sin θ$ , the polar form becomes: $z = r e^{i θ}$

Quadrant Adjustment for Argument

Quadrant	Condition	$θ$
I	$x > 0, y > 0$	$tan^{- 1} (\frac{y}{x})$
II	$x < 0, y > 0$	$\pi - \tan^{-1}!\left(\frac{
III	$x < 0, y < 0$	$-\pi + \tan^{-1}!\left(\frac{
IV	$x > 0, y < 0$	$-\tan^{-1}!\left(\frac{

Example: Convert $z = - 2 - 2 i$ to polar form.

$r = (- 2)^{2} + (- 2)^{2} = 8 = 22$
Reference angle: $α = tan^{- 1} (1) = \frac{π}{4}$
Since $z$ is in Quadrant III: $θ = - π + \frac{π}{4} = - \frac{3 π}{4}$
Polar form: $z = 22 (cos (- \frac{3 π}{4}) + i sin (- \frac{3 π}{4}))$

Operations with Complex Numbers in Polar Form

Multiplication

Division

$\frac{z _{1}}{z _{2}} = \frac{r _{1}}{r _{2}} e^{i (θ_{1} - θ_{2})}$ $\frac{z _{1}}{z _{2}} = \frac{∣ z _{1} ∣}{∣ z _{2} ∣}, ar g (\frac{z _{1}}{z _{2}}) = ar g (z_{1}) - ar g (z_{2})$

De Moivre's Theorem

$z^{n} = r^{n} (cos n θ + i sin n θ) = r^{n} e^{in θ}$

Modulus of a Product (Key Identity)

If $(x_{1} + i y_{1}) (x_{2} + i y_{2}) \dots (x_{n} + i y_{n}) = a + ib$ , then: $(x_{1}^{2} + y_{1}^{2}) (x_{2}^{2} + y_{2}^{2}) \dots (x_{n}^{2} + y_{n}^{2}) = a^{2} + b^{2}$

Equations and Identities in Polar Form

Using $a + b + c = 0$ to Prove Trigonometric Identities

If $cos α + cos β + cos γ = 0$ and $sin α + sin β + sin γ = 0$ , then:

Let $a = e^{i α}$ , $b = e^{i β}$ , $c = e^{iγ}$ . The conditions imply $a + b + c = 0$ .

Using the algebraic identity: $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$

Since $a + b + c = 0$ : $a^{3} + b^{3} + c^{3} = 3 ab c$

Substituting back: $e^{i 3 α} + e^{i 3 β} + e^{i 3 γ} = 3 e^{i (α + β + γ)}$

Taking real parts: $cos 3 α + cos 3 β + cos 3 γ = 3 cos (α + β + γ)$

Real-World Application: Complex Impedance

In AC circuit analysis, complex impedance $Z$ relates voltage $E$ and current $I$ : $Z = \frac{E}{I}$

To simplify division of complex numbers, multiply by the conjugate of the denominator: $Z = \frac{E}{I} = \frac{E \cdot I ˉ}{I \cdot I ˉ} = \frac{E I ˉ}{∣ I ∣ ^{2}}$