CIE AS/A Level Maths-3.9 Complex numbers- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-3.9 Complex numbers- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-3.9 Complex numbers- Study Notes
Key Concepts:
- Complex Numbers: Basic Ideas
- Operations on Complex Numbers (Cartesian form)
- Non-Real Roots Occur in Conjugate Pairs (Polynomials with Real Coefficients)
- Non-Real Roots Occur in Conjugate Pairs (Polynomials with Real Coefficients)
- Multiplication and Division in Polar (Exponential) Form
- Square Roots of Complex Numbers
- Geometrical Effects of Complex Number Operations
- Loci in an Argand Diagram
Complex Numbers: Basic Ideas
Complex Numbers
A complex number is a number of the form \(z = x + iy\), where \(x, y \in \mathbb{R}\) and \(i^2 = -1\).
Equality of Complex Numbers
- Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal.
- \(\text{Re } z\): Real part of \(z\).
- \(\text{Im } z\): Imaginary part of \(z\).
- \(|z|\): Modulus of \(z\), given by \(|z| = \sqrt{x^2 + y^2}\).
- \(\arg z\): Argument of \(z\), the angle \(\theta\) such that \(\tan \theta = \dfrac{y}{x}\).
- \(z^*\): Conjugate of \(z\), given by \(z^* = x – iy\).
Argument Convention
- The argument of a complex number usually refers to an angle \(\theta\) such that \(-\pi < \theta \leq \pi\).
- In some cases, it may be more convenient to use \(0 \leq \theta < 2\pi\).
- Unless specified, either interval may be used in answers.
Example:
For \(z = 1 – i\), find \(\text{Re } z\), \(\text{Im } z\), \(|z|\), \(\arg z\), and \(z^*\).
▶️ Answer/Explanation
Step 1: Real and Imaginary parts
\(\text{Re } z = 1,\; \text{Im } z = -1\).
Step 2: Modulus
\(|z| = \sqrt{1^2 + (-1)^2} = \sqrt{2}\).
Step 3: Argument
\(\tan \theta = \dfrac{-1}{1} = -1\).
The point \((1, -1)\) lies in the 4th quadrant, so \(\theta = -\dfrac{\pi}{4}\) (principal value).
Step 4: Conjugate
\(z^* = 1 + i\).
Final Answer: \(\text{Re } z = 1,\; \text{Im } z = -1,\; |z| = \sqrt{2},\; \arg z = -\dfrac{\pi}{4},\; z^* = 1 + i\).
Example:
Solve for \(x\) and \(y\) if \(z = x + iy\) satisfies \(z = 3 – 2i\).
▶️ Answer/Explanation
We use the fact that two complex numbers are equal if and only if both their real and imaginary parts are equal.
Step 1: Compare real parts
\(\text{Re } z = x = 3\).
Step 2: Compare imaginary parts
\(\text{Im } z = y = -2\).
Final Answer: \(\boxed{x = 3,\; y = -2}\).
Operations on Complex Numbers (Cartesian form)
Operations on Complex Numbers (Cartesian form)
Let \(z_1 = x_1 + i y_1\) and \(z_2 = x_2 + i y_2\), where \(x_1,y_1,x_2,y_2 \in \mathbb{R}\).
Addition
- \(z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)\).
Subtraction
- \(z_1 – z_2 = (x_1 – x_2) + i(y_1 – y_2)\).
Multiplication (full working)
Multiply by distributing and use \(i^2 = -1\):
- \(z_1 z_2 = (x_1 + i y_1)(x_2 + i y_2)\)
- \(= x_1 x_2 + i x_1 y_2 + i y_1 x_2 + i^2 y_1 y_2\)
- \(= (x_1 x_2 – y_1 y_2) + i(x_1 y_2 + x_2 y_1)\).
Division (full working)
Multiply numerator and denominator by the conjugate of the denominator to obtain a real denominator:
- \(\dfrac{z_1}{z_2} = \dfrac{x_1 + i y_1}{x_2 + i y_2} = \dfrac{(x_1 + i y_1)(x_2 – i y_2)}{(x_2 + i y_2)(x_2 – i y_2)}\).
- Denominator: \((x_2 + i y_2)(x_2 – i y_2) = x_2^2 + y_2^2\).
- Numerator: expand and simplify using \(i^2 = -1\) to get a complex number \(A + iB\).
- Final form: \(\dfrac{z_1}{z_2} = \dfrac{A}{x_2^2+y_2^2} + i\dfrac{B}{x_2^2+y_2^2}\), where \(A,B\) come from the expanded numerator.
- Equivalently: \(\dfrac{z_1}{z_2} = \dfrac{z_1 z_2^*}{|z_2|^2}\) where \(z_2^* = x_2 – i y_2\) and \(|z_2|^2 = x_2^2 + y_2^2\).
Inverse
- For \(z \neq 0\), \(z^{-1} = \dfrac{z^*}{|z|^2}\).
Example
Let \(z_1 = 3 + 2i\) and \(z_2 = -1 + 4i\). Find \(z_1 + z_2\) and \(z_1 – z_2\).
▶️ Answer/Explanation
Addition
\(z_1 + z_2 = (3 + (-1)) + i(2 + 4) = 2 + 6i\).
Subtraction
\(z_1 – z_2 = (3 – (-1)) + i(2 – 4) = 4 – 2i\).
Final Answers: \(\boxed{z_1 + z_2 = 2 + 6i,\; z_1 – z_2 = 4 – 2i}\).
Example
Compute \( (2 – 3i)(4 + i)\) showing full working.
▶️ Answer/Explanation
Expand and use \(i^2 = -1\).
\((2 – 3i)(4 + i) = 2\cdot 4 + 2\cdot i + (-3i)\cdot 4 + (-3i)\cdot i\)
\(= 8 + 2i – 12i – 3i^2\)
\(= 8 – 10i – 3(-1)\)
\(= 8 – 10i + 3 = 11 – 10i\).
Final Answer: \(\boxed{11 – 10i}\).
Example
Compute \(\dfrac{5 + 3i}{2 – i}\) showing full working.
▶️ Answer/Explanation
Multiply numerator and denominator by the conjugate of the denominator \(2 + i\).
\(\dfrac{5 + 3i}{2 – i} = \dfrac{(5 + 3i)(2 + i)}{(2 – i)(2 + i)}\).
Numerator: \((5 + 3i)(2 + i) = 5\cdot 2 + 5\cdot i + 3i\cdot 2 + 3i\cdot i\)
\(= 10 + 5i + 6i + 3i^2 = 10 + 11i + 3(-1)\)
\(= 7 + 11i\).
Denominator: \((2 – i)(2 + i) = 2^2 – (i)^2 = 4 – ( -1) = 5\).
Therefore \(\dfrac{5 + 3i}{2 – i} = \dfrac{7 + 11i}{5} = \dfrac{7}{5} + i\dfrac{11}{5}\).
Final Answer: \(\boxed{\dfrac{7}{5} + i\dfrac{11}{5}}\).
Non-Real Roots Occur in Conjugate Pairs (Polynomials with Real Coefficients)
Non-Real Roots Occur in Conjugate Pairs (Polynomials with Real Coefficients)
Statement
If a polynomial has real coefficients and a non-real complex number \(a + ib\) (with \(b \neq 0\)) is a root, then its complex conjugate \(a – ib\) is also a root. Thus non-real roots occur in conjugate pairs.
Method when one non-real root is given
- If a polynomial \(p(x)\) with real coefficients is known to have the root \(a + ib\), then \(a – ib\) is also a root.
- Form the quadratic factor coming from this conjugate pair: \(\bigl(x – (a+ib)\bigr)\bigl(x – (a-ib)\bigr) = (x-a)^2 + b^2\).
- Divide the given polynomial \(p(x)\) by the quadratic \((x-a)^2 + b^2\) to reduce the degree. The quotient will have real coefficients and can be solved by standard methods to find the remaining real or complex roots.
Example:
Given that \(x^3 – 7x^2 + 17x – 15 = 0\) has a root \(2 + i\), find all roots.
▶️ Answer/Explanation
Since coefficients are real and \(2 + i\) is a non-real root, its conjugate \(2 – i\) is also a root.
Step 1: Form the quadratic factor from the conjugate pair
\(\bigl(x – (2+i)\bigr)\bigl(x – (2-i)\bigr) = (x-2)^2 + 1 = x^2 – 4x + 5\).
Step 2: Divide the cubic by the quadratic to find the remaining linear factor
We look for \(x^3 – 7x^2 + 17x – 15 = (x^2 – 4x + 5)(x – r)\) for some real \(r\).
Expand the right hand side: \((x^2 – 4x + 5)(x – r) = x^3 – r x^2 – 4x^2 + 4r x + 5x – 5r\).
Collect like terms: \(= x^3 + (-r – 4)x^2 + (4r + 5)x – 5r\).
Equate coefficients with \(x^3 – 7x^2 + 17x – 15\):
\(-r – 4 = -7 \Rightarrow r = 3.\)
Check constant term: \(-5r = -15\) gives \(r = 3\) consistent.
Therefore the factorization is \(x^3 – 7x^2 + 17x – 15 = (x^2 – 4x + 5)(x – 3)\).
Step 3: Find all roots
From \(x – 3 = 0\) we get \(x = 3\).
From \(x^2 – 4x + 5 = 0\) use quadratic formula or complete the square: \(x = \dfrac{4 \pm \sqrt{(-4)^2 – 4\cdot 1\cdot 5}}{2} = \dfrac{4 \pm \sqrt{16 – 20}}{2} = \dfrac{4 \pm \sqrt{-4}}{2} = 2 \pm i.\)
Final Answer: \(\boxed{x = 3,\; x = 2 + i,\; x = 2 – i}\).
Example:
Given that \(x^4 – x^3 – 3x^2 + 17x – 30 = 0\) has a root \(1 + 2i\), find all roots.
▶️ Answer/Explanation
Because coefficients are real and \(1 + 2i\) is non-real, its conjugate \(1 – 2i\) must also be a root.
Step 1: Quadratic factor from the conjugate pair
\(\bigl(x – (1+2i)\bigr)\bigl(x – (1-2i)\bigr) = (x-1)^2 + (2)^2 = x^2 – 2x + 5\).
Step 2: Divide the quartic by \(x^2 – 2x + 5\)
Seek \(x^4 – x^3 – 3x^2 + 17x – 30 = (x^2 – 2x + 5)(x^2 + px + q)\).
Expand the right hand side and collect terms: \(= x^4 + p x^3 + q x^2 – 2x^3 – 2p x^2 – 2q x + 5x^2 + 5p x + 5q\).
Collecting powers gives \(= x^4 + (p – 2) x^3 + (q – 2p + 5) x^2 + (-2q + 5p) x + 5q\).
Equate coefficients with \(x^4 – x^3 – 3x^2 + 17x – 30\):
For \(x^3\): \(p – 2 = -1 \Rightarrow p = 1.\)
For \(x^2\): \(q – 2p + 5 = -3 \Rightarrow q – 2(1) + 5 = -3 \Rightarrow q + 3 = -3 \Rightarrow q = -6.\)
Check \(x\)-coefficient: \(-2q + 5p = -2(-6) + 5(1) = 12 + 5 = 17\) matches.
Check constant: \(5q = 5(-6) = -30\) matches.
Hence factorization: \(x^4 – x^3 – 3x^2 + 17x – 30 = (x^2 – 2x + 5)(x^2 + x – 6)\).
Step 3: Solve the quadratic factors
From \(x^2 – 2x + 5 = 0\) we get \(x = \dfrac{2 \pm \sqrt{4 – 20}}{2} = 1 \pm 2i\).
From \(x^2 + x – 6 = 0\) factor or use formula: \(x^2 + x – 6 = (x + 3)(x – 2)\) so \(x = -3\) or \(x = 2\).
Final Answer: \(\boxed{x = 1 + 2i,\; x = 1 – 2i,\; x = 2,\; x = -3}\).
Argand Diagram: Geometric Representation of Complex Numbers
Argand Diagram: Geometric Representation of Complex Numbers
An Argand diagram is the plane used to represent complex numbers graphically. A complex number \(z = x + iy\) corresponds to the point \((x,y)\) in the plane where the horizontal axis is the real axis and the vertical axis is the imaginary axis.
Plotting a Complex Number
- To plot \(z = x + iy\), mark the point with coordinates \((x,y)\).
- The real part \(\text{Re } z = x\) is the horizontal coordinate; the imaginary part \(\text{Im } z = y\) is the vertical coordinate.
Modulus as Distance
- The modulus \(|z|\) is the distance from the origin to the point \((x,y)\): \(|z| = \sqrt{x^2 + y^2}\).
- In geometric language \(|z|\) is the length of the vector from the origin to the point representing \(z\).
Argument as Angle
- The argument \(\arg z\) is the angle \(\theta\) measured from the positive real axis to the line joining the origin to \((x,y)\).
- It satisfies \(\tan \theta = \dfrac{\text{Im } z}{\text{Re } z} = \dfrac{y}{x}\), with quadrant determined by the signs of \(x\) and \(y\).
- The principal value is usually taken as \(-\pi < \theta \le \pi\), though \(0 \le \theta < 2\pi\) is sometimes used.
Geometric Effects of Algebraic Operations
- Multiplying by a positive real \(r\) scales distances from the origin by \(r\).
- Multiplying by \(e^{i\phi}\) rotates every point about the origin through angle \(\phi\) (counterclockwise if \(\phi>0\)).
- Conjugation \(z \mapsto z^*\) reflects the point \((x,y)\) across the real axis to \((x,-y)\).
Example:
Represent \(z = -2 + 2\sqrt{3}\,i\) on an Argand diagram and find \(\text{Re } z\), \(\text{Im } z\), \(|z|\), and \(\arg z\) (principal value).
▶️ Answer/Explanation
Step 1: Coordinates
\(\text{Re } z = -2,\; \text{Im } z = 2\sqrt{3}\). Plot the point \((-2,\;2\sqrt{3})\) with the real axis horizontal and imaginary axis vertical.
Step 2: Modulus
\(|z| = \sqrt{(-2)^2 + \bigl(2\sqrt{3}\bigr)^2} = \sqrt{4 + 12} = \sqrt{16} = 4\).
Step 3: Argument
\(\tan \theta = \dfrac{2\sqrt{3}}{-2} = -\sqrt{3}\).
The point \((-2,2\sqrt{3})\) lies in the second quadrant where \(\tan \theta = -\sqrt{3}\) corresponds to reference angle \(\dfrac{\pi}{3}\). Thus the principal argument is \(\theta = \pi – \dfrac{\pi}{3} = \dfrac{2\pi}{3}\).
Step 4: Geometric description
On the Argand diagram the point lies 4 units from the origin at angle \(\dfrac{2\pi}{3}\) from the positive real axis.
Final Answer: \(\text{Re } z = -2,\; \text{Im } z = 2\sqrt{3},\; |z| = 4,\; \arg z = \dfrac{2\pi}{3}.\)
Multiplication and Division in Polar (Exponential) Form
Multiplication and Division in Polar (Exponential) Form
Polar / Exponential Form
A complex number may be written as \(z = r(\cos \theta + i\sin \theta) = r e^{i\theta}\), where \(r = |z| \ge 0\) and \(\theta = \arg z\).
Multiplication
- If \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\) then \( z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}. \)
- Consequently \(|z_1 z_2| = |z_1|\,|z_2| = r_1 r_2\) and \(\arg(z_1 z_2) = \arg z_1 + \arg z_2\) (taken modulo \(2\pi\) or choosing principal values as required).
Division
- If \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\) with \(r_2 \neq 0\) then \( \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\; e^{i(\theta_1 – \theta_2)}. \)
- Consequently \(\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|} = \dfrac{r_1}{r_2}\) and \(\arg\!\left(\dfrac{z_1}{z_2}\right) = \arg z_1 – \arg z_2\) (modulo \(2\pi\) or using principal values as required).
Example
Let \(z_1 = 2 e^{i\pi/6}\) and \(z_2 = 3 e^{i\pi/4}\). Compute \(z_1 z_2\) and give modulus and argument.
▶️ Answer/Explanation
Step 1: Multiply moduli
\(|z_1 z_2| = r_1 r_2 = 2 \cdot 3 = 6\).
Step 2: Add arguments
\(\arg(z_1 z_2) = \theta_1 + \theta_2 = \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{2\pi + 3\pi}{12} = \dfrac{5\pi}{12}.\)
Step 3: Write product in polar/exponential form
\(z_1 z_2 = 6\, e^{i\frac{5\pi}{12}} = 6\bigl(\cos \tfrac{5\pi}{12} + i\sin \tfrac{5\pi}{12}\bigr).\)
Optional: Rectangular form
\(z_1 z_2 = 6\cos\!\tfrac{5\pi}{12} + i\,6\sin\!\tfrac{5\pi}{12}\) (evaluate numerically if required).
Final Answer: \(\boxed{z_1 z_2 = 6\, e^{i\frac{5\pi}{12}},\; |z_1 z_2| = 6,\; \arg(z_1 z_2) = \dfrac{5\pi}{12}.}\)
Example
Let \(z_1 = 5 e^{i\frac{3\pi}{4}}\) and \(z_2 = 2 e^{-i\frac{\pi}{6}}\). Compute \(\dfrac{z_1}{z_2}\) and give modulus and argument (principal value chosen where convenient).
▶️ Answer/Explanation
Step 1: Divide moduli
\(\left|\dfrac{z_1}{z_2}\right| = \dfrac{r_1}{r_2} = \dfrac{5}{2}.\)
Step 2: Subtract arguments
\(\arg\!\left(\dfrac{z_1}{z_2}\right) = \theta_1 – \theta_2 = \dfrac{3\pi}{4} -\bigl(-\dfrac{\pi}{6}\bigr) = \dfrac{3\pi}{4} + \dfrac{\pi}{6} = \dfrac{9\pi + 2\pi}{12} = \dfrac{11\pi}{12}.\)
Step 3: Write quotient in polar/exponential form
\(\dfrac{z_1}{z_2} = \dfrac{5}{2}\; e^{i\frac{11\pi}{12}} = \dfrac{5}{2}\bigl(\cos \tfrac{11\pi}{12} + i\sin \tfrac{11\pi}{12}\bigr).\)
Note on principal argument
\(\dfrac{11\pi}{12}\) lies in \((0,\pi)\) so it is already a principal value in \(-\pi < \theta \le \pi\).
Final Answer: \(\boxed{\dfrac{z_1}{z_2} = \dfrac{5}{2}\, e^{i\frac{11\pi}{12}},\; \left|\dfrac{z_1}{z_2}\right| = \dfrac{5}{2},\; \arg\!\left(\dfrac{z_1}{z_2}\right) = \dfrac{11\pi}{12}.}\)
Square Roots of Complex Numbers
Square Roots of Complex Numbers
The square roots of a complex number \(z\) are the complex numbers \(w\) such that \(w^2 = z\). Every nonzero complex number has exactly two square roots, which are negatives of each other.
Method 1 (Algebraic: Cartesian Form)
- Suppose \(z = a + ib\) and let its square root be \(w = x + iy\), with \(x, y \in \mathbb{R}\).
- Then \((x + iy)^2 = x^2 – y^2 + i(2xy) = a + ib\).
- Equating real and imaginary parts gives the system:
- \(x^2 – y^2 = a\)
- \(2xy = b\)
- From this system we can solve for \(x\) and \(y\).
Method 2 (Polar/Exponential Form)
- Write the number in polar form: \(z = r\big(\cos \theta + i\sin \theta\big) = re^{i\theta}\), where \(r = |z|\) and \(\theta = \arg(z)\).
- The square roots are given by: \[ w = \sqrt{r}\, e^{i\theta/2}, \quad \text{and} \quad w = \sqrt{r}\, e^{i(\theta/2 + \pi)}. \]
- Thus the two square roots differ by a minus sign.
Important Notes
- Every nonzero complex number has exactly two square roots.
- If \(z\) is purely real and positive, the roots are \(\pm \sqrt{z}\) (real numbers).
- If \(z\) is purely real and negative, the roots are purely imaginary numbers \(\pm i\sqrt{|z|}\).
- For non-real \(z\), both roots will be complex numbers in Cartesian form.
Example:
Find the two square roots of \(5 + 12i\) in exact Cartesian form. Show full working.
▶️ Answer/Explanation
Let the square root be \(z = x + iy\).
Then \((x + iy)^2 = x^2 – y^2 + i(2xy)\). Equating with \(5 + 12i\): \( x^2 – y^2 = 5, \quad 2xy = 12. \)
From \(2xy = 12\), we get \(xy = 6\).
Thus \(y = \dfrac{6}{x}\).
Substitute into the real-part equation.
\(x^2 – \left(\dfrac{6}{x}\right)^2 = 5 \;\;\Rightarrow\;\; x^4 – 36 = 5x^2\).
Rearrange: \(x^4 – 5x^2 – 36 = 0\).
Let \(u = x^2\). Then \(u^2 – 5u – 36 = 0\).
Solve quadratic in \(u\).
Discriminant = \(25 + 144 = 169\), \(\sqrt{169} = 13\).
So \(u = \dfrac{5 \pm 13}{2}\).
Hence \(u = 9\) or \(u = -4\).
Since \(u = x^2 \geq 0\), we take \(u = 9\). So \(x = \pm 3\).
Find corresponding \(y\).
If \(x = 3\), then \(y = \dfrac{6}{3} = 2\).
If \(x = -3\), then \(y = \dfrac{6}{-3} = -2\).
Final Answer: \(\boxed{3 + 2i \quad \text{and} \quad -3 – 2i}\).
Geometrical Effects of Complex Number Operations
Geometrical Effects of Complex Number Operations
Conjugation
- If \(z = x + iy\), then its conjugate is \(z^* = x – iy\).
- On the Argand diagram, conjugation reflects the point across the real axis.
- So the point \((x, y)\) becomes \((x, -y)\).
Addition and Subtraction
- Adding two complex numbers corresponds to adding their vectors on the Argand diagram (parallelogram rule).
- Subtracting corresponds to vector subtraction (the difference between position vectors).
Geometrically, this is exactly like addition and subtraction of 2D vectors.
Multiplication
If \(z_1 = r_1(\cos\theta_1 + i\sin\theta_1)\) and \(z_2 = r_2(\cos\theta_2 + i\sin\theta_2)\), then \( z_1 z_2 = r_1 r_2 \big( \cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2) \big). \)
Effect:
- The modulus is multiplied: \(|z_1 z_2| = |z_1||z_2|\).
- The arguments are added: \(\arg(z_1 z_2) = \arg(z_1) + \arg(z_2)\).
Geometrically: scaling (by modulus) and rotation (by argument).
Division
If \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\), then \[ \dfrac{z_1}{z_2} = \dfrac{r_1}{r_2} e^{i(\theta_1 – \theta_2)}. \]
Effect:
- The modulus is divided: \(\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}\).
- The arguments are subtracted: \(\arg\left(\dfrac{z_1}{z_2}\right) = \arg(z_1) – \arg(z_2)\).
Geometrically: shrinking/stretching and rotation in the opposite direction.
Example:
Let \(z_1 = 1 + i\) and \(z_2 = \sqrt{3} + i\). Describe geometrical effects of conjugation, multiplication, and division.
▶️ Answer/Explanation
Conjugate of \(z_1 = 1 + i\) is \(z_1^* = 1 – i\). On Argand diagram, this is reflection across the real axis.
First write in polar form: \(|z_1| = \sqrt{2}, \; \arg(z_1) = \dfrac{\pi}{4}\). \(|z_2| = 2, \; \arg(z_2) = \dfrac{\pi}{6}\).
Product: \(|z_1 z_2| = \sqrt{2} \times 2 = 2\sqrt{2}\), Argument = \(\dfrac{\pi}{4} + \dfrac{\pi}{6} = \dfrac{5\pi}{12}\).
So multiplication corresponds to rotation by \(\dfrac{5\pi}{12}\) and scaling by \(2\sqrt{2}\).
Quotient modulus: \(\dfrac{|z_1|}{|z_2|} = \dfrac{\sqrt{2}}{2}\). Argument: \(\dfrac{\pi}{4} – \dfrac{\pi}{6} = \dfrac{\pi}{12}\).
So division corresponds to shrinking by a factor \(\dfrac{\sqrt{2}}{2}\) and rotating by \(\dfrac{\pi}{12}\).
Loci in an Argand Diagram
Loci in an Argand Diagram
Circles and Discs
- The condition \(|z – a| = k\) represents the locus of points whose distance from the fixed point \(a\) is exactly \(k\). Geometrically: a circle centered at \(a\) with radius \(k\).
- The condition \(|z – a| < k\) represents the set of points inside this circle (open disc). Geometrically: the interior region of the circle.
- The condition \(|z – a| \leq k\) includes the circle boundary.
Perpendicular Bisectors
- The condition \(|z – a| = |z – b|\) represents all points equidistant from \(a\) and \(b\).
- Geometrically: this is the perpendicular bisector of the line segment joining \(a\) and \(b\).
Half-Lines (Rays)
- The condition \(\arg(z – a) = \alpha\) represents all points \(z\) such that the line from \(a\) to \(z\) makes an angle \(\alpha\) with the positive real axis.
- Geometrically: this is a half-line (ray) starting from \(a\) at angle \(\alpha\).
- If inequality is used, e.g. \(\arg(z – a) < \alpha\), then the locus is a region between rays (a sector).
Example:
Describe and sketch the loci represented by the following equations:
- \(|z – (2 + i)| < 3\)
- \(|z – 1| = |z + 3|\)
- \(\arg(z – i) = \dfrac{\pi}{4}\)
▶️ Answer/Explanation
1. \(|z – (2 + i)| < 3\)
This is the interior of a circle centered at \((2,1)\) with radius \(3\).
2. \(|z – 1| = |z + 3|\)
Points equidistant from \(1 + 0i\) and \(-3 + 0i\). This is the perpendicular bisector of the line joining \((-3,0)\) and \((1,0)\).
Midpoint = \((-1,0)\), so the locus is the vertical line \(x = -1\).
3. \(\arg(z – i) = \dfrac{\pi}{4}\)
Points \(z\) such that the line from \((0,1)\) to \(z\) makes an angle of \(\pi/4\) with the positive real axis.
This is a half-line (ray) starting at \((0,1)\) and going out at \(45^\circ\).