Complex Numbers
HL 2020 – Question 2 (Complex Numbers)
25 marks(a) Find the two complex numbers \(z_1\) and \(z_2\) that satisfy the following simultaneous equations, where \(i^2 = -1\):
\(i z_1 = -4 + 3i\)
\(3z_1 - z_2 = 11 + 17i\)
Write your answers in the form \(a + bi\), where \(a, b \in \mathbb{Z}\).
Solution (a)
From \(i z_1 = -4 + 3i\), multiply both sides by \(i\):
\(i^2 z_1 = i(-4 + 3i)\)
\(-z_1 = -4i + 3i^2 = -4i - 3\)
Hence \(z_1 = 3 + 4i\).
Substitute into \(3z_1 - z_2 = 11 + 17i\):
\(3(3 + 4i) - z_2 = 11 + 17i\)
\(9 + 12i - z_2 = 11 + 17i\)
Rearranging gives
\(-z_2 = 11 + 17i - 9 - 12i = 2 + 5i\)
so \(z_2 = -2 - 5i\). Therefore \(z_1 = 3 + 4i\) and \(z_2 = -2 - 5i\).
Low partial (2): Any correct step involving multiplying or dividing one equation by \(i\), e.g. writing \(i(-4 + 3i)\).
Mid partial (3): One of \(z_1\) or \(z_2\) correctly found, or an equation correctly formed with one variable eliminated but not fully solved.
High partial (4): A correct elimination step leading to one of \(z_1 = 3 + 4i\) or \(z_2 = -2 - 5i\), but the second value not fully obtained or clearly stated.
Full credit (5): Both values correctly obtained and clearly stated: \(z_1 = 3 + 4i\) and \(z_2 = -2 - 5i\).
(b) The complex numbers \(3 + 2i\) and \(5 - i\) are the first two terms of a geometric sequence.
(i) Find \(r\), the common ratio of the sequence. Write your answer in the form \(a + bi\), where \(a, b \in \mathbb{Z}\).
Solution (b)(i)
The first term is \(T_1 = 3 + 2i\) and the second is \(T_2 = 5 - i\).
\(r = \dfrac{T_2}{T_1} = \dfrac{5 - i}{3 + 2i}\)
Multiply numerator and denominator by \(3 - 2i\):
\(r = \dfrac{(5 - i)(3 - 2i)}{(3 + 2i)(3 - 2i)}\)
\((5 - i)(3 - 2i) = 15 - 10i - 3i + 2i^2 = 13 - 13i\), and \((3 + 2i)(3 - 2i) = 9 - 4i^2 = 13\).
\(r = \dfrac{13 - 13i}{13} = 1 - i\)
Low partial (3): Sets up the correct quotient \(\dfrac{5 - i}{3 + 2i}\) or begins multiplying by the conjugate.
High partial (4): Most of the algebra correct, with at most one slip.
Full credit (5): Completes the calculation correctly and gives \(r = 1 - i\).
(b)(ii) Use de Moivre’s Theorem to find \(T_9\), the ninth term of the sequence. Write your answer in the form \(a + bi\), where \(a, b \in \mathbb{Z}\).
Solution (b)(ii)
For a geometric sequence \(T_n = a r^{\,n-1}\) with \(a = T_1 = 3 + 2i\) and \(r = 1 - i\),
\(T_9 = a r^8 = (3 + 2i)(1 - i)^8\)
For \(1 - i\), \(|1 - i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\) and a suitable argument is \(-\tfrac{\pi}{4}\), so
\(1 - i = \sqrt{2}\bigl(\cos(-\tfrac{\pi}{4}) + i\sin(-\tfrac{\pi}{4})\bigr)\)
By de Moivre’s Theorem,
\((1 - i)^8 = (\sqrt{2})^8\bigl(\cos(-2\pi) + i\sin(-2\pi)\bigr) = 16\)
\(T_9 = (3 + 2i)\cdot16 = 48 + 32i\)
so the ninth term is \(T_9 = 48 + 32i\).
Low partial (4): Any correct use of de Moivre’s Theorem in context, e.g. writing \(1 - i\) in polar form or attempting \((1 - i)^8\); or a correct use of \(T_n = a r^{\,n-1}\) with complex \(a, r\).
Mid partial (7): Correct modulus and argument for \(1 - i\), or a largely correct de Moivre calculation with some errors.
High partial (11): A correct polar-form approach with most working correct, but a slip in simplifying \((1 - i)^8\) or in multiplying by \(3 + 2i\).
Full credit (15): Complete correct solution giving \(T_9 = 48 + 32i\). (If a different value of \(r\) from part (b)(i) is used consistently, mark on that work.)
Complex Numbers — Selected Questions
1. Equality of complex numbers
Find the complex number \(z=x+yi\) if \(5z+2i\overline{z}=11-4i\), where \(\overline{z}\) is the conjugate of \(z\).
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2. Addition, subtraction and multiplication
If \(z=5-3i\) and \(w=-2+4i\), express in the form \(a+bi\): \[ 3w\big(2z-\overline{z}\big). \]
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3. Conjugate and division
Express \(\displaystyle \frac{30+i}{3+5i}\) in the form \(a+bi\).
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4. Square roots
Find real \(a,b\) if \((a+bi)^2=3-4i\).
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5. Argand diagram and modulus
Let \(z=3+4i\) and \(w=1-2i\). Compute \(z+w\) and investigate whether \(|z+w|=|z|+|w|\).
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6. Interpreting an Argand diagram
The Argand diagram below shows points \(a,b,c,d,e,f\).
They represent \(z,\;2z,\;\overline{z},\;z+\overline{z},\;iz,\; (1+i)z\) .
(i) Identify the label for each of \(z,\;2z,\;\overline{z},\;iz,\;z+\overline{z},\;(1+i)z\).
(ii) Determine the acute angle z, O, (1+i)z.
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Identification rules.
- \(2z\): same argument as \(z\); modulus doubled (lies on same ray, twice as far).
- \(\overline{z}\): reflection of \(z\) in the real axis.
- \(iz\): rotation of \(z\) by \(+90^\circ\) (same modulus).
- \(z+\overline{z}\): real part is doubled, imaginary parts add to 0.
- \((1+i)z\): multiply by \(1+i=\sqrt2\,\mathrm{cis}(\pi/4)\) → scale by \(\sqrt2\) and rotate \(+45^\circ\).
Angle at \(O\). Multiplication by \(1+i\) rotates any vector by \(\arg(1+i)=45^\circ\). Thus \(\angle z, O,\,(1+i)z=\boxed{45^\circ}\) for any \(z\neq 0\).
| Complex number | Label on diagram |
|---|---|
| \(z\) | c |
| \(2z\) | e |
| \(\overline{z}\) | b |
| \(iz\) | a |
| \(z+\overline{z}\) | f |
| \((1+i)z\) | d |
7. Complex equations
If \(2+i\) is a root of \(z^2-(a-i)z+(8+bi)=0\) with \(a,b\in\mathbb{R}\), find \(a,b\) and the other root.
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8. Conjugate Roots Theorem
Suppose \(3+2i\) is a root of \(z^3+az^2+bz-52=0\) with \(a,b\in\mathbb{R}\). Find \(a,b\) and the other roots.
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9. Polar form
Express \(-2\sqrt3+2i\) in the form \(r(\cos\alpha+i\sin\alpha)\).
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10. De Moivre: Trigonometric identity
Using De Moivre’s Theorem, express \(\sin 3\theta\) as a polynomial in \(\sin\theta\).
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11. De Moivre: Large powers
Express \(\big(1-\sqrt3\,i\big)^8\) in the form \(a+bi\).
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12. Roots via De Moivre
Solve \(z^3=-64\) and give the roots in \(a+bi\) form.
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1. Irrational numbers
Irrational numbers - surds |
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Construct root 2Using a compass and straight edge
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Construct root 3Using a compass and straight edge
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2. Complex numbers introduction
Equation with imaginary roots |
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Quadratic equation with imaginary rootsUsing the quadratic formula
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Addition and multiplication of complex numbers |
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3. Division and equity of complex numbers
Dividing by a complex number - conjugateExpressing a fraction with a real denominator
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Complex number equations"Real = real, imaginary = imaginary"
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4. Argand diagram - modulus
Plot on an Argand diagram
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5. Transformations of complex numbers
Understanding multiplication |
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Multiplication involves scaling and rotating
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Transformations involving multiplication |
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6. Conjugate roots theorem
Quadratic equation - conjugate roots |
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Showing that a conjugate is a root |
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Solving a cubic equation with two imaginary roots |
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7. Polar form of a complex number
Converting from polar to rectangular form |
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Express in polar formFinding the modulus and argument
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8. Products and quotients in polar form
Multiplying and dividing in polar formMultiplication: Multiply the module and add the arguments
Division: Divide the module and subtract the arguments
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9. De Moivre's Theorem
Raising a number to a power - expanding |
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Express in polar form and expand |
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10. Applications of de Moivre's Theorem
Raising a number to a power |
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Proving trigonometric identities using de Moivre's Theorem |
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Finding roots |
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Solving Complex Equations:
Real parts = Real parts Imaginary parts = Imaginary parts |
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Transformations in the Complex Plane
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Complex conjugate roots
Solving quadratic and cubic equations with imaginary roots. |
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Writing a complex number in Polar Form
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Multiplying and dividing in Polar Form and using deMoivre's Theorem to expand a complex number to a power.
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Using deMoivre's Theorem to find roots of a Complex Equation.
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Using deMoivre's Theorem to prove Trigonometric identities. This question also requires use of the Binomial Theorem.
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The Syllabus
Abraham de Moivre (1667–1754)
Abraham de Moivre was a French mathematician best known for De Moivre’s Theorem, which links complex numbers and trigonometry through powers of \(\cos\theta + i\sin\theta\). Forced to leave France, he lived most of his life in London and became associated with Newton and Halley. His work in probability and complex numbers remains foundational in modern mathematics.
De Moivre’s Theorem
Induction proof for positive integers \(n\)Theorem. For \(n \ge 1\) and real \(\theta\),
\[ (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta). \]
Base case \(n = 1\). We have \((\cos\theta + i\sin\theta)^1 = \cos\theta + i\sin\theta = \cos(1\theta) + i\sin(1\theta)\), so the result holds for \(n=1\).
Induction hypothesis. Assume for some integer \(k \ge 1\) that
\[ (\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta). \tag{1} \]
Induction step. Then
\[ (\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)(\cos(k\theta) + i\sin(k\theta)). \]
Expanding and using \(i^2=-1\),
\[ \begin{aligned} (\cos\theta + i\sin\theta)^{k+1} &= \cos\theta\cos(k\theta) - \sin\theta\sin(k\theta) \\ &\quad {}+ i\bigl(\cos\theta\sin(k\theta) + \sin\theta\cos(k\theta)\bigr). \end{aligned} \]
By the angle–addition formulae, \(\cos(a+b)=\cos a\cos b-\sin a\sin b\) and \(\sin(a+b)=\sin a\cos b+\cos a\sin b\), with \(a=\theta\), \(b=k\theta\), this is
\[ \cos((k+1)\theta) + i\sin((k+1)\theta). \]
Conclusion. The theorem holds for \(n=1\), and if it holds for \(n=k\) it holds for \(n=k+1\). By induction, De Moivre’s theorem is true for all integers \(n \ge 1\).
