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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

AS & A Level Maths Study Notes– All Topics

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.

Key Terms and Notations

  • \(\text{Re } z\): 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:

  1. \(|z – (2 + i)| < 3\)
  2. \(|z – 1| = |z + 3|\)
  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 \(45^\circ\).

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