IB Mathematics AA Complex roots & de Moivre’s theorem Study Notes
IB Mathematics AA Complex roots & de Moivre’s theorem Study Notes
IB Mathematics AA Complex roots & de Moivre’s theorem Study Notes Offer a clear explanation of Complex roots & de Moivre’s theorem , 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 Complex roots & de Moivre’s theorem
Complex Roots and De Moivre’s Theorem
Complex roots and De Moivre’s theorem are critical concepts in advanced mathematics. They provide insights into polynomial equations with real coefficients and enable calculations of powers and roots of complex numbers efficiently.
Key Concepts
Complex Conjugate Roots:
- If a polynomial equation has real coefficients, any non-real roots must occur in conjugate pairs.
- Example: For a polynomial equation \( f(x) = x^3 – 4x + 13 = 0 \), if \( z = a + bi \) is a root, then \( \overline{z} = a – bi \) is also a root.
- De Moivre’s Theorem:
- For \( z = r\text{cis}\theta \), where \( n \in \mathbb{Z}^+ \), the theorem states:
- \( z^n = r^n\text{cis}(n\theta) \).
- Extended to rational exponents:
- \( z^{1/n} = r^{1/n}\text{cis}\left(\frac{\theta + 2k\pi}{n}\right), k = 0, 1, \ldots, n-1 \).
- For \( z = r\text{cis}\theta \), where \( n \in \mathbb{Z}^+ \), the theorem states:
- Powers and Roots of Complex Numbers:
- Link to the sum and product of roots of polynomial equations (AHL 2.12).
- Connection to compound angle identities (AHL 3.10).
Guidance, Clarifications, and Syllabus Links
- Complex conjugates simplify calculations for polynomials with real coefficients.
- De Moivre’s theorem has practical applications in trigonometry and root-finding for polynomials.
- The geometric interpretation of De Moivre’s theorem relates to rotations and dilations in the complex plane.
Examples
Example 1: Complex Conjugate Roots
Given \( f(x) = x^2 – 2x + 5 = 0 \), find the roots:
Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \).
Here, \( a = 1, b = -2, c = 5 \): \( x = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i \).
The roots are \( 1 + 2i \) and \( 1 – 2i \), which are conjugates.
Example 2: Application of De Moivre’s Theorem
Find \( (1 + i)^6 \):
Convert to polar form: \( z = \sqrt{2}\text{cis}\frac{\pi}{4} \).
Using De Moivre’s theorem: \( z^6 = (\sqrt{2})^6\text{cis}(6 \cdot \frac{\pi}{4}) = 8\text{cis}\frac{3\pi}{2} \).
Convert back to Cartesian form: \( z = -8i \).
Example 3: Roots of Unity
Find the cube roots of \( z = 8 \):
Convert 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 Complex roots & de Moivre’s theorem Exam Style Worked Out Questions
Question
Consider the equation \(\frac{2z}{3-z}\) = i where z = x + i y and x , y ∈ R.
Find the value of x and the value of y . [Maximum mark: 5]
▶️Answer/Explanation
Ans:
substituting z = x + iy and *z = x – iy
\(\frac{2(x+iy)}{3-(x-iy)}\)=i
2x+2iy = -y + i (3-x)
equate real and imaginary:
y= -2x AND 2y= 3-x
solving simultaneously: x= -1, y=2 (z=-1+2i)
Question
Let \(z = \cos \theta + i\sin \theta \).
a. Use de Moivre’s theorem to find the value of \({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3}\).[2]
▶️Answer/Explanation
Ans:
\({\left( {\cos \left( {\frac{\pi }{3}} \right) + {\text{i}}\sin \left( {\frac{\pi }{3}} \right)} \right)^3} = \cos \pi + {\text{i}}\sin \pi \) M1
\( = – 1\) A1
[2 marks]
b. Use mathematical induction to prove that
\[{(\cos \theta – {\text{i}}\sin \theta )^n} = \cos n\theta – {\text{i}}\sin n\theta {\text{ for }}n \in {\mathbb{Z}^ + }.\][6]
▶️Answer/Explanation
Ans:
show the expression is true for \(n = 1\) R1
assume true for \(n = k,{\text{ }}{(\cos \theta – {\text{i}}\sin \theta )^k} = \cos k\theta – {\text{i}}\sin k\theta \) M1
Note: Do not accept “let \(n = k\)” or “assume \(n = k\)”, assumption of truth must be present.
\({(\cos \theta – {\text{i}}\sin \theta )^{k + 1}} = {(\cos \theta – {\text{i}}\sin \theta )^k}(\cos \theta – {\text{i}}\sin \theta )\)
\( = (\cos k\theta – {\text{i}}\sin k\theta )(\cos \theta – {\text{i}}\sin \theta )\) M1
\( = \cos k\theta \cos \theta – \sin k\theta \sin \theta – {\text{i}}(\cos k\theta \sin \theta + \sin k\theta \cos \theta )\) A1
Note: Award A1 for any correct expansion.
\( = \cos \left( {(k + 1)\theta } \right) – {\text{i}}\sin \left( {(k + 1)\theta } \right)\) A1
therefore if true for \(n = k\) true for \(n = k + 1\), true for \(n = 1\), so true for all \(n( \in {\mathbb{Z}^ + })\) R1
Note: To award the final R mark the first 4 marks must be awarded.
[6 marks]
c. Find an expression in terms of \(\theta \) for \({(z)^n} + {(z{\text{*}})^n},{\text{ }}n \in {\mathbb{Z}^ + }\) where \(z{\text{*}}\) is the complex conjugate of \(z\).[2]
▶️Answer/Explanation
Ans:
\({(z)^n} + {(z{\text{*}})^n} = {(\cos \theta + {\text{i}}\sin \theta )^n} + {(\cos \theta – {\text{i}}\sin \theta )^n}\)
\( = \cos n\theta + {\text{i}}\sin n\theta + \cos n\theta – {\text{i}}\sin n\theta = 2\cos (n\theta )\) (M1)A1
[2 marks]
▶️Answer/Explanation
Ans:
\(zz* = (\cos \theta + {\text{i}}\sin \theta )(\cos \theta – {\text{i}}\sin \theta )\)
\( = {\cos ^2}\theta + {\sin ^2}\theta \) A1
\( = 1\) AG
Note: Allow justification starting with \(|z| = 1\).
(ii) Write down the binomial expansion of \({(z + z{\text{*}})^3}\) in terms of \(z\) and \(z{\text{*}}\).
▶️Answer/Explanation
Ans: \({(z + z{\text{*}})^3} = {z^3} + 3{z^2}z{\text{*}} + 3z{({z^*})^2} + (z{\text{*}})3\left( { = {z^3} + 3z + 3z{\text{*}} + {{(z{\text{*}})}^3}} \right)\) A1
(iii) Hence show that \(\cos 3\theta = 4{\cos ^3}\theta – 3\cos \theta \).[5]
▶️Answer/Explanation
Ans:
\({(z + z{\text{*}})^3} = {(2\cos \theta )^3}\) A1
\({z^3} + 3z + 3z{\text{*}} + {(z{\text{*}})^3} = 2\cos 3\theta + 6\cos \theta \) M1A1
\(\cos 3\theta = 4{\cos ^3}\theta – 3\cos \theta \) AG
Note: M1 is for using \(zz{\text{*}} = 1\), this might be seen in d(ii).
[5 marks]
e. Hence solve \(4{\cos ^3}\theta – 2{\cos ^2}\theta – 3\cos \theta + 1 = 0\) for \(0 \leqslant \theta < \pi \).[6]
▶️Answer/Explanation
Ans:
\(4{\cos ^3}\theta – 2{\cos ^2}\theta – 3\cos \theta + 1 = 0\)
\(4{\cos ^3}\theta – 3\cos \theta = 2{\cos ^2}\theta – 1\)
\(\cos (3\theta ) = \cos (2\theta )\) A1A1
Note: A1 for \(\cos (3\theta )\) and A1 for \(\cos (2\theta )\).
\(\theta = 0\) A1
or \(3\theta = 2\pi – 2\theta {\text{ }}({\text{or }}3\theta = 4\pi – 2\theta )\) M1
\(\theta = \frac{{2\pi }}{5},{\text{ }}\frac{{4\pi }}{5}\) A1A1
Note: Do not accept solutions via factor theorem or other methods that do not follow “hence”.
[6 marks]