IB Mathematics AA HL Polar Forms Euler's Form Study Notes
IB Mathematics AA HL Polar Forms Euler’s Form Study Notes
IB Mathematics AA HL Polar Forms Euler’s Form Study Notes Offer a clear explanation of Polar Forms Euler’s Form , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Polar Forms Euler’s Form.
Polar Form, Euler Form, and De Moivre’s Theorem
Complex numbers can be expressed in various forms, including Cartesian form, polar form, and Euler form. These representations allow for efficient computation of sums, products, quotients, powers, and roots of complex numbers.
Key Concepts
- Modulus–Argument (Polar) Form:
- The polar form of a complex number is \( z = r(\cos\theta + i\sin\theta) \), also written as \( z = r\text{cis}\theta \), where:
- \( r = |z| = \sqrt{a^2 + b^2} \): modulus (magnitude).
- \( \theta = \text{arg}(z) = \tan^{-1}\left(\frac{b}{a}\right) \): argument (angle).
- The polar form of a complex number is \( z = r(\cos\theta + i\sin\theta) \), also written as \( z = r\text{cis}\theta \), where:
- Euler Form:
- Using Euler’s formula, \( e^{i\theta} = \cos\theta + i\sin\theta \), the polar form can be rewritten as:
- \( z = re^{i\theta} \).
- Using Euler’s formula, \( e^{i\theta} = \cos\theta + i\sin\theta \), the polar form can be rewritten as:
- Conversions Between Forms:
- From Cartesian to Polar/Euler:
- Find \( r = \sqrt{a^2 + b^2} \) and \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).
- Express as \( z = r\text{cis}\theta \) or \( z = re^{i\theta} \).
- From Polar/Euler to Cartesian:
- Expand \( z = r(\cos\theta + i\sin\theta) \) or \( z = re^{i\theta} \) to get \( z = a + bi \).
- From Cartesian to Polar/Euler:
- Sums, Products, and Quotients:
- Geometric interpretation is important for understanding operations on complex numbers in polar or Euler form:
- Product: \( z_1 \cdot z_2 = r_1r_2\text{cis}(\theta_1 + \theta_2) \).
- Quotient: \( \frac{z_1}{z_2} = \frac{r_1}{r_2}\text{cis}(\theta_1 – \theta_2) \).
- Geometric interpretation is important for understanding operations on complex numbers in polar or Euler form:
- Complex Conjugates:
- Complex conjugate roots occur in pairs for polynomials with real coefficients.
- De Moivre’s Theorem:
- For \( z = r\text{cis}\theta \) and \( n \in \mathbb{Z}^+ \), \( z^n = r^n\text{cis}(n\theta) \).
- Extension to rational exponents: \( z^{1/n} = r^{1/n}\text{cis}\left(\frac{\theta + 2k\pi}{n}\right), k = 0, 1, \ldots, n-1 \).
Guidance, Clarifications:
- Proof of De Moivre’s theorem for \( n \in \mathbb{Z}^+ \) is expected using mathematical induction.
- Applications include:
- Finding powers and roots of complex numbers.
- Link to compound angle identities: \( (\cos\theta + i\sin\theta)^n \) expands to trigonometric expressions.
- Connections to polynomial equations:
- Roots of a polynomial can be found using \( z^{1/n} \), and the sum and product of roots relate to coefficients (AHL 2.12).
Examples
Example 1: Convert \( z = 1 + \sqrt{3}i \) to Polar Form
- Modulus: \( r = \sqrt{1^2 + (\sqrt{3})^2} = 2 \).
- Argument: \( \theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \).
- Polar Form: \( z = 2\text{cis}\frac{\pi}{3} \) or \( z = 2e^{i\pi/3} \).
Example 2: Use De Moivre’s Theorem to Find \( (1 + i)^4 \)
- Convert to polar form: \( z = \sqrt{2}\text{cis}\frac{\pi}{4} \).
- Using De Moivre’s theorem: \( z^4 = (\sqrt{2})^4\text{cis}(4 \cdot \frac{\pi}{4}) = 4\text{cis}\pi = -4 \).
Example 3: Roots of \( z^3 = 8 \)
- Convert \( z \) to polar form: \( z = 8\text{cis}0 \).
- Roots: \( z^{1/3} = 2\text{cis}\left(\frac{0 + 2k\pi}{3}\right), k = 0, 1, 2 \).
- Roots: \( z_0 = 2, z_1 = 2\text{cis}\frac{2\pi}{3}, z_2 = 2\text{cis}\frac{4\pi}{3} \).
IB Mathematics AA SL Polar Forms Euler's Form Exam Style Worked Out Questions
[Maximum mark: 18]
Question:
In the following Argand diagram, the points Z1 , O and Z2 are the vertices of triangle Z1OZ2 described anticlockwise.
The point Z1 represents the complex number z1 = r1eiα , where r1 > 0. The point Z2 represents the complex number z2 = r2eiθ , where r2 > 0.
Angles α, θ are measured anticlockwise from the positive direction of the real axis such that 0 ≤ α, θ < 2π and 0 < α – θ < π.
(a) Show that z1z2 * = r1r2e i(α – θ) where z2* is the complex conjugate of z2 .
▶️Answer/Explanation
Ans:
Note: Accept working in modulus-argument form
(b) Given that Re (z1 z2*) = 0, show that Z1OZ2 is a right-angled triangle.
In parts (c), (d) and (e), consider the case where Z1OZ2 is an equilateral triangle.
▶️Answer/Explanation
Ans:
(c) (i) Express z1 in terms of z2 .
▶️Answer/Explanation
Ans:
Note: Accept working in either modulus-argument form to obtain \(z_{1} = z_{2}\left ( cos\frac{\pi }{3} + isin\frac{\pi }{3} \right )\) or in Cartesian form to obtain \(z_{1} = z_{2}\left ( \frac{1}{2} + \frac{\sqrt{3}}{2} \right )i\)
(ii) Hence show that z12 + z22 = z1 z2 .
Let z1 and z2 be the distinct roots of the equation z2 + az + b = 0 where z ∈ C and a , b ∈ R.
▶️Answer/Explanation
Ans:
(d) Use the result from part (c)(ii) to show that a2 – 3b = 0.
Consider the equation z2 + az + 12 = 0, where z ∈ C and a ∈ R.
▶️Answer/Explanation
Ans:
(e) Given that 0 < α – θ < π, deduce that only one equilateral triangle Z1OZ2 can be formed from the point O and the roots of this equation.
▶️Answer/Explanation
Ans:
so (for 0 < α – θ < π), only one equilateral triangle can be formed from point O and the two roots of this equation
Question
Consider \(w = 2\left( {{\text{cos}}\frac{\pi }{3} + {\text{i}}\,{\text{sin}}\frac{\pi }{3}} \right)\)
a. These four points form the vertices of a quadrilateral, Q.
i. Express w2 and w3 in modulus-argument form.[3]
▶️Answer/Explanation
Ans: \({w^2} = 4\text{cis}\left( {\frac{{2\pi }}{3}} \right){\text{;}}\,\,{w^3} = 8{\text{cis}}\left( \pi \right)\) (M1)A1A1
Note: Accept Euler form.
Note: M1 can be awarded for either both correct moduli or both correct arguments.
Note: Allow multiplication of correct Cartesian form for M1, final answers must be in modulus-argument form.
[3 marks]
ii. Sketch on an Argand diagram the points represented by w0 , w1 , w2 and w3.[2]
▶️Answer/Explanation
Ans:
A1A1
[2 marks]
b. Show that the area of the quadrilateral Q is \(\frac{{21\sqrt 3 }}{2}\).[3]
▶️Answer/Explanation
Ans:
use of area = \(\frac{1}{2}ab\,\,{\text{sin}}\,C\) M1
\(\frac{1}{2} \times 1 \times 2 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 2 \times 4 \times {\text{sin}}\frac{\pi }{3} + \frac{1}{2} \times 4 \times 8 \times {\text{sin}}\frac{\pi }{3}\) A1A1
Note: Award A1 for \(C = \frac{\pi }{3}\), A1 for correct moduli.
\( = \frac{{21\sqrt 3 }}{2}\) AG
Note: Other methods of splitting the area may receive full marks.
[3 marks]
Let \(z = 2\left( {{\text{cos}}\frac{\pi }{n} + {\text{i}}\,{\text{sin}}\frac{\pi }{n}} \right),\,\,n \in {\mathbb{Z}^ + }\). The points represented on an Argand diagram by \({z^0},\,\,{z^1},\,\,{z^2},\, \ldots \,,\,\,{z^n}\) form the vertices of a polygon \({P_n}\).
c. Show that the area of the polygon \({P_n}\) can be expressed in the form \(a\left( {{b^n} – 1} \right){\text{sin}}\frac{\pi }{n}\), where \(a,\,\,b\, \in \mathbb{R}\).[6]
▶️Answer/Explanation
Ans:
\(\frac{1}{2} \times {2^0} \times {2^1} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^1} \times {2^2} \times {\text{sin}}\frac{\pi }{n} + \frac{1}{2} \times {2^2} \times {2^3} \times {\text{sin}}\frac{\pi }{n} + \, \ldots \, + \frac{1}{2} \times {2^{n – 1}} \times {2^n} \times {\text{sin}}\frac{\pi }{n}\) M1A1
Note: Award M1 for powers of 2, A1 for any correct expression including both the first and last term.
\( = {\text{sin}}\frac{\pi }{n} \times \left( {{2^0} + {2^2} + {2^4} + \, \ldots \, + {2^{n – 2}}} \right)\)
identifying a geometric series with common ratio 22(= 4) (M1)A1
\( = \frac{{1 – {2^{2n}}}}{{1 – 4}} \times {\text{sin}}\frac{\pi }{n}\) M1
Note: Award M1 for use of formula for sum of geometric series.
\( = \frac{1}{3}\left( {{4^n} – 1} \right){\text{sin}}\frac{\pi }{n}\) A1
[6 marks]