IB Mathematics AI SL Modulus–argument (polar) form Study Notes - New Syllabus

♦THE POLAR FORM (MODULUS-ARGUMENT FORM)
A complex number \( z = x + yi \) can also be described using polar coordinates \((r, \theta)\):
\( r = \) length of the vector (modulus \( |z| \)).
\( \theta = \) angle between the \( x \)-axis and the vector (argument \( \arg(z) \)).

♦Relations:
\( \cos \theta = \frac{x}{r}, \quad \sin \theta = \frac{y}{r}, \quad \tan \theta = \frac{y}{x} \)

Thus, \( z \) can be written in polar form:
\( z = r (\cos \theta + i \sin \theta) \)

♦REMARK:
The argument \( \theta \) is not unique. For the principal argument, we agree:
\( -180^\circ < \theta \leq 180^\circ \quad \text{or} \quad -\pi < \theta \leq \pi \)

♦Transformation from Cartesian to Polar Form:
Given \( z = x + yi \), find \( r \) and \( \theta \):
\( r = |z| = \sqrt{x^2 + y^2} \)
\( \tan \theta = \frac{y}{x}, \quad \text{considering the quadrant of } (x, y). \)

Example Find the polar form of \( z = 1 + \sqrt{3}i \) and \( w = 3 + 4i \). ▶️Answer/Explanation Solution:  For \( z = 1 + \sqrt{3}i \): \( r = \sqrt{1 + 3} = 2 \), \( \tan \theta = \sqrt{3} \) (1st quadrant) \( \Rightarrow \theta = \frac{\pi}{3} \). Thus, \( z = 2 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) \). For \( w = 3 + 4i \): \( r = 5 \), \( \tan \theta = \frac{4}{3} \) (1st quadrant) \( \Rightarrow \theta \approx 0.927 \) (by GDC). Thus, \( w = 5 \left( \cos 0.927 + i \sin 0.927 \right) \).

Example Find the polar form of: \( z_1 = 1 + i : ? \) \( z_2 = -1 – i : ? \) \( z_3 = 1 – i : ? \) \( z_4 = -1 + i : ? \) ▶️Answer/Explanation Solution: \( z_1 = 1 + i \): \( \theta = \frac{\pi}{4} \) \( z_2 = -1 – i \): \( \theta = \frac{5\pi}{4} \) \( z_3 = 1 – i \): \( \theta = -\frac{\pi}{4} \) \( z_4 = -1 + i \): \( \theta = \frac{3\pi}{4} \) All have modulus \( \sqrt{2} \).

♦Transformation from Polar to Cartesian Form:
Simply compute \( x = r \cos \theta \) and \( y = r \sin \theta \).

GDC Tip: Use the “COMPLEX” mode to switch between forms.

♦CIS FORM AND EULER’S FORM
CIS Form: \( z = r \operatorname{cis} \theta = r (\cos \theta + i \sin \theta) \).
Euler’s Form: \( z = r e^{i\theta} \), where \( e^{i\theta} = \cos \theta + i \sin \theta \).

Example All forms of \( z_1 = 1 + i \), \( z_2 = 3 + 4i \), \( z_3 = 3 – 4i \): ▶️Answer/Explanation Solution:

♦NOTICE
Any complex number with modulus 1 has polar form \( z = \operatorname{cis} \theta \).
 For real numbers \( \pm a \), the argument is \( 0 \) (if \( a > 0 \)) or \( \pi \) (if \( a < 0 \)).
 For imaginary numbers \( \pm ai \), the argument is \( \frac{\pi}{2} \) or \( -\frac{\pi}{2} \).
 The conjugate of \( z = r (\cos \theta + i \sin \theta) \) is \( \overline{z} = r (\cos \theta – i \sin \theta) = r \operatorname{cis}(-\theta) \).

♦ PRODUCTS, QUOTIENTS, AND POWERS IN POLAR FORM
For \( z_1 = r_1 \operatorname{cis} \theta_1 \) and \( z_2 = r_2 \operatorname{cis} \theta_2 \):
Product: \( z_1 z_2 = r_1 r_2 \operatorname{cis}(\theta_1 + \theta_2) \).
Quotient: \( \frac{z_1}{z_2} = \frac{r_1}{r_2} \operatorname{cis}(\theta_1 – \theta_2) \).
Power (De Moivre’s Law): \( z^n = r^n \operatorname{cis}(n \theta) \).

♦Properties:
 Modulus: \( |z_1 z_2| = |z_1| |z_2| \), \( \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} \), \( |z^n| = |z|^n \).
 Argument: \( \arg(z_1 z_2) = \arg(z_1) + \arg(z_2) \), \( \arg\left( \frac{z_1}{z_2} \right) = \arg(z_1) – \arg(z_2) \), \( \arg(z^n) = n \arg(z) \).

Example Let \( z = 2 \operatorname{cis} \frac{\pi}{6} \) and \( w = \operatorname{cis} \frac{\pi}{3} \).  \( z w=?\)  \( \frac{z}{w}=?\)  \( z^6=?\) ▶️Answer/Explanation Solution:  \( z w = 2 \operatorname{cis} \frac{\pi}{2} = 2i \).  \( \frac{z}{w} = 2 \operatorname{cis}\left( -\frac{\pi}{6} \right) = \sqrt{3} – i \).  \( z^6 = 64 \operatorname{cis} \pi = -64 \).

Example Calculate \( (1 + i)^{10} \): ▶️Answer/Explanation Solution:  \( 1 + i = \sqrt{2} \operatorname{cis} \frac{\pi}{4} \).  \( (1 + i)^{10} = (\sqrt{2})^{10} \operatorname{cis} \frac{10\pi}{4} = 32 \operatorname{cis} \frac{\pi}{2} = 32i \).

Example Using De Moivre’s theorem, derive trigonometric identities:  \( z = \operatorname{cis} \theta \Rightarrow z^2 = \cos 2\theta + i \sin 2\theta \). ▶️Answer/Explanation Solution:  \( z = \operatorname{cis} \theta \Rightarrow z^2 = \cos 2\theta + i \sin 2\theta \).  Also, \( z^2 = (\cos \theta + i \sin \theta)^2 = (\cos^2 \theta – \sin^2 \theta) + i (2 \sin \theta \cos \theta) \).  Comparing gives: \( \cos 2\theta = \cos^2 \theta – \sin^2 \theta \), \( \sin 2\theta = 2 \sin \theta \cos \theta \).

♦ GEOMETRICAL INTERPRETATION OF MULTIPLICATION
Multiplying by \( z = k \) scales the vector by \( k \).
Multiplying by \( z = \operatorname{cis} \phi \) rotates the vector by \( \phi \).
Multiplying by \( z = k \operatorname{cis} \phi \) scales and rotates the vector.

♦ ADDING SINUSOIDAL FUNCTIONS
Express sums of sinusoidal functions as a single function:
\( A_1 \cos(x + \theta_1) + A_2 \cos(x + \theta_2) = A \cos(x + \theta) \).
Use Euler’s form: \( A \cos(x + \theta) = \operatorname{Re}(A e^{i(x + \theta)}) \).

Example For \( f(x) = 4 \cos x \) and \( g(x) = 3 \cos\left(x + \frac{\pi}{4}\right) \): ▶️Answer/Explanation Solution: \( f(x) + g(x) = \operatorname{Re}(4 e^{ix} + 3 e^{i(x + \pi/4)}) = \operatorname{Re}(6.48 e^{i(x + 0.334)}) = 6.48 \cos(x + 0.334) \). Application: Adding waves of the same frequency results in a wave of the same frequency.

Example For \( V_1(t) = 7 \sin(4t – 1) \) and \( V_2(t) = 2 \sin(4t + 3) \) ▶️Answer/Explanation Solution:  \( V_1(t) + V_2(t) = 5.89 \sin(4t – 1.26) \).