Figure 1.2: The Argand Plane, Modulus-Argument & Complex Conjugate
A complex number $z = a + bi$ plotted as a position vector $(a, b)$ with modulus $r = \sqrt{a^2+b^2}$ and principal argument $\theta \in (-\pi, \pi]$. Its complex conjugate $z^* = a - bi$ is the reflection across the real axis with argument $-\theta$.
1. The Complex Number System & Cartesian Form
The field of complex numbers $\mathbb{C}$ extends the real numbers $\mathbb{R}$ by introducing the imaginary unit $i$, defined such that $i^2 = -1$ (or $i = \sqrt{-1}$). Every complex number $z \in \mathbb{C}$ can be uniquely written in Cartesian (rectangular) form:
$$z = a + bi \quad (a, b \in \mathbb{R})$$- Real Part: $\text{Re}(z) = a \in \mathbb{R}$
- Imaginary Part: $\text{Im}(z) = b \in \mathbb{R}$ (Crucial: $\text{Im}(z)$ is the real coefficient $b$, NOT $bi$!).
- Powers of $i$ Cyclic Property: $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$. In general, $i^{4k+r} = i^r$ for remainder $r \in \{0, 1, 2, 3\}$.
2. Complex Conjugation & Division
The complex conjugate of $z = a + bi$ is defined as $z^* = a - bi$ (also written $\bar{z}$). Geometrically, $z^*$ is the reflection of $z$ across the real axis $\text{Re}(z)$.
- Sum & Difference: $z + z^* = 2\text{Re}(z) = 2a \in \mathbb{R}$, and $z - z^* = 2i\text{Im}(z) = 2bi \in i\mathbb{R}$.
- Modulus Squared Identity: $z z^* = (a + bi)(a - bi) = a^2 - b^2 i^2 = a^2 + b^2 = |z|^2 \in \mathbb{R}^+$.
- Cartesian Division Strategy: To divide $\frac{w}{z}$, multiply numerator and denominator by the conjugate $z^*$ to make the denominator strictly real: $$\frac{w}{z} = \frac{w \cdot z^*}{z \cdot z^*} = \frac{w z^*}{|z|^2}$$
3. Modulus, Principal Argument & Polar Form
A. Modulus ($r$ or $|z|$)
The modulus represents the absolute Euclidean distance from the origin $(0, 0)$ to the point $(a, b)$ in the Argand plane:
$$r = |z| = \sqrt{a^2 + b^2} \ge 0$$Triangle Inequality: $|z_1 + z_2| \le |z_1| + |z_2|$ and $||z_1| - |z_2|| \le |z_1 - z_2|$.
B. The Principal Argument ($\theta = \arg(z)$)
Under the IB DP Mathematics AA HL curriculum, the argument is strictly required to lie in the principal range:
$$-\pi < \theta \le \pi \quad (\text{or } -180^\circ < \theta \le 180^\circ)$$To compute $\theta$ without quadrant error, find the acute reference angle $\alpha = \arctan\left|\frac{b}{a}\right|$ and determine the sign branch by quadrant:
| Quadrant | Signs of $(a, b)$ | Principal Argument Formula | Example |
|---|---|---|---|
| Quadrant 1 | $a > 0, b > 0$ | $\theta = \alpha$ | $1 + i\sqrt{3} \implies \theta = \frac{\pi}{3}$ |
| Quadrant 2 | $a < 0, b > 0$ | $\theta = \pi - \alpha$ | $-1 + i \implies \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ |
| Quadrant 3 | $a < 0, b < 0$ | $\theta = -(\pi - \alpha) = -\pi + \alpha$ | $-\sqrt{3} - i \implies \theta = -\left(\pi - \frac{\pi}{6}\right) = -\frac{5\pi}{6}$ |
| Quadrant 4 | $a > 0, b < 0$ | $\theta = -\alpha$ | $1 - i \implies \theta = -\frac{\pi}{4}$ |
C. Polar Form: $z = r(\cos\theta + i\sin\theta) \equiv r\text{ cis}\theta$
Converting Cartesian coordinates via $a = r\cos\theta$ and $b = r\sin\theta$ produces the polar representation.
4. Euler’s Exponential Form & Geometric Operations
By Maclaurin series expansion, Euler established the fundamental identity connecting the complex exponential and trigonometric functions:
$$e^{i\theta} = \cos\theta + i\sin\theta \quad (\theta \in \mathbb{R})$$This yields the Euler exponential form: $z = r e^{i\theta}$.
- Multiplication (Rotation & Scaling): $$z_1 z_2 = (r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = (r_1 r_2)e^{i(\theta_1 + \theta_2)}$$ Rule: Multiply the moduli ($|z_1 z_2| = |z_1||z_2|$) and add the arguments ($\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) \pmod{2\pi}$).
- Division: $$\frac{z_1}{z_2} = \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} = \left(\frac{r_1}{r_2}\right)e^{i(\theta_1 - \theta_2)}$$ Rule: Divide the moduli ($|\frac{z_1}{z_2}| = \frac{|z_1|}{|z_2|}$) and subtract the arguments ($\arg(\frac{z_1}{z_2}) = \arg(z_1) - \arg(z_2) \pmod{2\pi}$).
- Euler’s Trigonometric Identities: $$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
5. Geometric Loci in the Complex Plane
In the IB DP AA HL exam, equations and inequalities in $z$ describe geometric figures in the Argand plane:
| Complex Locus Equation | Geometric Interpretation | Cartesian Conversion ($z = x + iy$) |
|---|---|---|
| $|z - z_0| = R$ | Circle centered at $z_0 = (x_0, y_0)$ with radius $R$ | $(x - x_0)^2 + (y - y_0)^2 = R^2$ |
| $|z - z_1| = |z - z_2|$ | Perpendicular bisector of line segment joining $z_1$ and $z_2$ | Straight line $Ax + By + C = 0$ |
| $\arg(z - z_0) = \alpha$ | Half-line (ray) originating at $z_0$ making angle $\alpha$ with positive real axis (excluding $z_0$) | $y - y_0 = \tan(\alpha)(x - x_0)$ with directional constraint |
| $\text{Re}(z) = c$ or $\text{Im}(z) = k$ | Vertical line $x = c$ or horizontal line $y = k$ | $x = c$ or $y = k$ |
Worked Examples
Worked Example 1: Simultaneous Complex Equation with Conjugates (Paper 1 Standard)
Find the complex number $z = x + iy$ ($x, y \in \mathbb{R}$) that satisfies the equation: $$3z - 2i z^* = 7 + 9i$$
Solution:
1. Substitute $z = x + iy$ and $z^* = x - iy$: [M1]
$$3(x + iy) - 2i(x - iy) = 7 + 9i$$2. Expand and simplify powers of $i$: [M1]
$$3x + 3iy - 2ix + 2i^2 y = 7 + 9i$$ $$\text{Since } i^2 = -1, \quad 2i^2 y = -2y$$ $$(3x - 2y) + (3y - 2x)i = 7 + 9i \quad \text{[A1]}$$3. Equate real and imaginary parts to form a system of simultaneous linear equations: [M1]
$$\begin{cases} 3x - 2y = 7 & \text{(1)} \\ -2x + 3y = 9 & \text{(2)} \end{cases}$$4. Solve by elimination: multiply (1) by $3$ and (2) by $2$: [M1]
$$9x - 6y = 21$$ $$-4x + 6y = 18$$ $$\text{Adding equations: } 5x = 39 \implies x = \frac{39}{5} = 7.8$$Substitute $x = \frac{39}{5}$ into (1):
$$3\left(\frac{39}{5}\right) - 2y = 7 \implies \frac{117}{5} - 7 = 2y \implies \frac{82}{5} = 2y \implies y = \frac{41}{5} = 8.2 \quad \text{[A1]}$$5. Assemble the complex number: [A1]
$$z = \frac{39}{5} + \frac{41}{5}i \quad (\text{or } 7.8 + 8.2i)$$Worked Example 2: Conversion to Euler Form & Angle Normalization (Paper 1 Standard)
Let $z_1 = -1 + i\sqrt{3}$ and $z_2 = 1 - i$.
(a) Express $z_1$ and $z_2$ in the Euler exponential form $r e^{i\theta}$, where $-\pi < \theta \le \pi$. [4 marks]
(b) Hence express $\frac{z_1}{z_2}$ in Euler form with principal argument. [2 marks]
(c) By converting $\frac{z_1}{z_2}$ to Cartesian form, deduce the exact value of $\cos\left(\frac{11\pi}{12}\right)$. [3 marks]
Solution:
(a) Polar/Euler Conversions:
For $z_1 = -1 + i\sqrt{3}$: [M1]
$r_1 = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2$.
$a < 0, b > 0 \implies \text{Quadrant 2}$. Reference angle $\alpha_1 = \arctan(\sqrt{3}) = \frac{\pi}{3}$.
$\theta_1 = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
For $z_2 = 1 - i$: [M1]
$r_2 = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.
$a > 0, b < 0 \implies \text{Quadrant 4}$. Reference angle $\alpha_2 = \arctan(1) = \frac{\pi}{4}$.
$\theta_2 = -\frac{\pi}{4}$.
(b) Quotient Evaluation:
$$\frac{z_1}{z_2} = \frac{2e^{i\frac{2\pi}{3}}}{\sqrt{2}e^{-i\frac{\pi}{4}}} = \frac{2}{\sqrt{2}} e^{i\left(\frac{2\pi}{3} - \left(-\frac{\pi}{4}\right)\right)} = \sqrt{2} e^{i\left(\frac{8\pi + 3\pi}{12}\right)} = \sqrt{2}e^{i\frac{11\pi}{12}} \quad \text{[M1][A1]}$$Notice $-\pi < \frac{11\pi}{12} \le \pi$, so the argument is already in principal range.
(c) Cartesian Equating & Exact Trig Value:
Calculate $\frac{z_1}{z_2}$ directly in Cartesian form by realizing the denominator: [M1]
$$\frac{-1 + i\sqrt{3}}{1 - i} = \frac{(-1 + i\sqrt{3})(1 + i)}{(1 - i)(1 + i)} = \frac{-1 - i + i\sqrt{3} + i^2\sqrt{3}}{1^2 + 1^2} = \frac{(-1 - \sqrt{3}) + (\sqrt{3} - 1)i}{2}$$Equate the real parts: $\text{Re}\left(\sqrt{2}e^{i\frac{11\pi}{12}}\right) = \sqrt{2}\cos\left(\frac{11\pi}{12}\right)$: [M1]
$$\sqrt{2}\cos\left(\frac{11\pi}{12}\right) = \frac{-1 - \sqrt{3}}{2} \implies \cos\left(\frac{11\pi}{12}\right) = \frac{-1 - \sqrt{3}}{2\sqrt{2}} = -\frac{\sqrt{6} + \sqrt{2}}{4} \quad \text{[A1]}$$Worked Example 3: Complex Loci Intersection (Paper 2 Standard)
In the complex plane, let $L_1$ be the locus of points satisfying $|z - 3| = 2$, and let $L_2$ be the locus satisfying $|z - 1| = |z - 5i|$.
(a) Describe geometrically $L_1$ and $L_2$. [2 marks]
(b) Find the Cartesian equations of $L_1$ and $L_2$. [3 marks]
(c) Determine the exact coordinates of the points of intersection of $L_1$ and $L_2$. [4 marks]
Solution:
(a) Geometric Descriptions:
$L_1$: Circle with center $(3, 0)$ and radius $R = 2$. [A1]
$L_2$: Perpendicular bisector of the line segment joining $(1, 0)$ and $(0, 5)$. [A1]
(b) Cartesian Equations:
For $L_1$: $|(x - 3) + iy| = 2 \implies (x - 3)^2 + y^2 = 4$. [A1]
For $L_2$: $|(x - 1) + iy|^2 = |x + i(y - 5)|^2$: [M1]
(c) Intersection Points:
Substitute $y = \frac{x + 12}{5}$ into $(x - 3)^2 + y^2 = 4$: [M1]
$$(x - 3)^2 + \left(\frac{x + 12}{5}\right)^2 = 4 \implies (x^2 - 6x + 9) + \frac{x^2 + 24x + 144}{25} = 4$$Multiply through by $25$: [M1]
$$25(x^2 - 6x + 9) + (x^2 + 24x + 144) = 100$$ $$25x^2 - 150x + 225 + x^2 + 24x + 144 = 100$$ $$26x^2 - 126x + 269 = 0$$Evaluate discriminant: $\Delta = (-126)^2 - 4(26)(269) = 15876 - 27976 = -12100 < 0$. [R1]
Since $\Delta < 0$, there are no real solutions $(x, y)$, meaning $L_1$ and $L_2$ do not intersect. [A1]
Worked Example 4: Euler Trigonometric Synthesis Identity (Paper 1 Standard)
Use the exponential identity $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ to prove that: $$\sin^3\theta = \frac{3\sin\theta - \sin(3\theta)}{4}$$
Solution:
1. Cube both sides of Euler's sine formula: [M1]
$$\sin^3\theta = \left( \frac{e^{i\theta} - e^{-i\theta}}{2i} \right)^3 = \frac{(e^{i\theta} - e^{-i\theta})^3}{(2i)^3}$$2. Evaluate the denominator: $(2i)^3 = 8 i^3 = 8(-i) = -8i$. [A1]
3. Expand the numerator using the binomial theorem $(a - b)^3 = a^3 - 3a^2 b + 3a b^2 - b^3$: [M1]
$$(e^{i\theta} - e^{-i\theta})^3 = (e^{i\theta})^3 - 3(e^{i\theta})^2(e^{-i\theta}) + 3(e^{i\theta})(e^{-i\theta})^2 - (e^{-i\theta})^3$$ $$= e^{3i\theta} - 3e^{i\theta} + 3e^{-i\theta} - e^{-3i\theta} \quad \text{[A1]}$$4. Group terms into conjugate pairs: [M1]
$$= (e^{3i\theta} - e^{-3i\theta}) - 3(e^{i\theta} - e^{-i\theta})$$5. Substitute back into the expression over $-8i$: [M1]
$$\sin^3\theta = \frac{(e^{3i\theta} - e^{-3i\theta}) - 3(e^{i\theta} - e^{-i\theta})}{-8i} = \frac{3(e^{i\theta} - e^{-i\theta}) - (e^{3i\theta} - e^{-3i\theta})}{8i}$$ $$= \frac{3}{4}\left( \frac{e^{i\theta} - e^{-i\theta}}{2i} \right) - \frac{1}{4}\left( \frac{e^{3i\theta} - e^{-3i\theta}}{2i} \right) \quad \text{[A1]}$$6. Recognize the definitions $\sin\theta$ and $\sin(3\theta)$: [R1]
$$\sin^3\theta = \frac{3}{4}\sin\theta - \frac{1}{4}\sin(3\theta) = \frac{3\sin\theta - \sin(3\theta)}{4} \quad \text{[AG]}$$Worked Example 5: Geometric Locus in the Argand Plane (Paper 1 Standard)
The locus of a complex number $z = x + iy$ satisfies the equation $|z - 3i| = |z - (4 + i)|$.
(a) Give a geometric description of this locus. [1 mark]
(b) Find the Cartesian equation of the locus in the form $y = mx + c$. [4 marks]
Solution:
(a) Geometric Description:
The equation $|z - z_1| = |z - z_2|$ represents the perpendicular bisector of the line segment joining the points $A(0, 3)$ and $B(4, 1)$ in the complex plane [R1].
(b) Algebraic Derivation:
Let $z = x + iy$:
$$|x + i(y - 3)| = |(x - 4) + i(y - 1)| \quad [M1]$$Square both sides:
$$x^2 + (y - 3)^2 = (x - 4)^2 + (y - 1)^2 \quad [M1]$$Expand both sides:
$$x^2 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 2y + 1$$Cancel $x^2 + y^2$ and simplify:
$$-6y + 9 = -8x - 2y + 17 \quad [A1]$$ $$-4y = -8x + 8 \implies 4y = 8x - 8 \implies y = 2x - 2 \quad [A1]$$Geometric Check: Midpoint of $A(0,3)$ and $B(4,1)$ is $M(2, 2)$. Slope of $AB = \frac{1-3}{4-0} = -\frac{1}{2}$. Perpendicular slope $= 2$. Line: $y - 2 = 2(x - 2) \implies y = 2x - 2$. Verified!
Practice Exercises (10 Tiered Questions)
Complete these 10 exam-style questions spanning core arithmetic, Cartesian-polar-Euler conversions, complex loci, and trigonometric synthesis. Open each question to inspect the full step-by-step IB markscheme.
Express $\frac{5 - i}{2 + 3i}$ in the form $a + bi$, where $a, b \in \mathbb{R}$.
Show Full Solution
Multiply numerator and denominator by conjugate of denominator ($2 - 3i$): [M1]
$$\frac{5 - i}{2 + 3i} \times \frac{2 - 3i}{2 - 3i} = \frac{10 - 15i - 2i + 3i^2}{2^2 + 3^2} \quad \text{[M1]}$$Substitute $i^2 = -1$: [A1]
$$\frac{10 - 3 - 17i}{4 + 9} = \frac{7 - 17i}{13} = \frac{7}{13} - \frac{17}{13}i \quad \text{[A1]}$$Solve the equation $z^2 - 4z + 13 = 0$, giving your answers in the form $a \pm bi$.
Show Full Solution
Apply quadratic formula: [M1]
$$z = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(13)}}{2(1)} = \frac{4 \pm \sqrt{16 - 52}}{2} \quad \text{[A1]}$$Simplify radical: $\sqrt{-36} = 6i$: [A1]
$$z = \frac{4 \pm 6i}{2} = 2 \pm 3i \implies z = 2 + 3i \quad \text{or} \quad z = 2 - 3i \quad \text{[A1]}$$Find the modulus and principal argument of $z = 1 - i\sqrt{3}$, and express $z$ in polar form $r\text{ cis}\theta$.
Show Full Solution
Modulus: $r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2$. [A1]
Argument: $a > 0, b < 0 \implies \text{Quadrant 4}$. [M1]
$$\theta = -\arctan\left(\frac{\sqrt{3}}{1}\right) = -\frac{\pi}{3} \quad \text{[A1]}$$ $$z = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right) = 2\text{ cis}\left(-\frac{\pi}{3}\right) \quad \text{[A1]}$$Given $w_1 = 6e^{i\frac{\pi}{4}}$ and $w_2 = 3e^{i\frac{5\pi}{6}}$, find $\frac{w_1}{w_2}$ in Euler form $r e^{i\theta}$ where $-\pi < \theta \le \pi$.
Show Full Solution
Divide moduli: $r = \frac{6}{3} = 2$. [A1]
Subtract arguments: [M1]
$$\theta = \frac{\pi}{4} - \frac{5\pi}{6} = \frac{3\pi - 10\pi}{12} = -\frac{7\pi}{12} \quad \text{[A1]}$$Check principal range: $-\pi < -\frac{7\pi}{12} \le \pi$. [R1]
$$\frac{w_1}{w_2} = 2e^{-i\frac{7\pi}{12}} \quad \text{[A1]}$$Find the complex number $z = x + iy$ that satisfies $2z + 3z^* = 15 - 4i$.
Show Full Solution
Substitute $z = x + iy$ and $z^* = x - iy$: [M1]
$$2(x + iy) + 3(x - iy) = 5x - iy = 15 - 4i \quad \text{[A1]}$$Equate real and imaginary parts: [M1]
$$5x = 15 \implies x = 3, \quad -y = -4 \implies y = 4 \quad \text{[A1]}$$ $$z = 3 + 4i \quad \text{[A1]}$$The locus of points $z$ in the complex plane satisfies $|z - 2| = |z - 4i|$. Find the Cartesian equation of this locus in terms of $x$ and $y$.
Show Full Solution
Substitute $z = x + iy$ and square both sides: [M1]
$$|(x - 2) + iy|^2 = |x + i(y - 4)|^2 \implies (x - 2)^2 + y^2 = x^2 + (y - 4)^2 \quad \text{[A1]}$$ $$x^2 - 4x + 4 + y^2 = x^2 + y^2 - 8y + 16 \quad \text{[M1]}$$ $$-4x + 4 = -8y + 16 \implies 8y = 4x + 12 \implies y = \frac{1}{2}x + \frac{3}{2} \quad \text{[A1]}$$Geometric Note: This is the perpendicular bisector of the segment between $(2, 0)$ and $(0, 4)$.
Describe geometrically the locus in the complex plane given by $|z - (3 - 2i)| = 5$, and write down its Cartesian equation.
Show Full Solution
Form $|z - z_0| = R$ represents a circle: [R1]
Center: $z_0 = 3 - 2i \implies (3, -2)$; Radius: $R = 5$. [A1]
Cartesian equation: [A1]
$$(x - 3)^2 + (y + 2)^2 = 25$$Given $z = x + iy$, find the geometric condition on $x$ and $y$ such that $\frac{z - 1}{z + 1}$ is purely imaginary.
Show Full Solution
Substitute $z = x + iy$: $\frac{(x - 1) + iy}{(x + 1) + iy}$. [M1]
Multiply by conjugate $[(x + 1) - iy]$: [M1]
$$\text{Numerator } = [(x - 1) + iy][(x + 1) - iy] = (x^2 - 1 + y^2) + i(y(x + 1) - y(x - 1))$$For the quotient to be purely imaginary, $\text{Re}(\text{Numerator}) = 0$: [R1]
$$x^2 + y^2 - 1 = 0 \implies x^2 + y^2 = 1 \quad (\text{with } (x, y) \neq (-1, 0)) \quad \text{[A1]}$$Conclusion: The locus is the unit circle centered at the origin, punctured at $(-1, 0)$.
Using Euler's relations $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$ and $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$, prove that $\cos^2\theta = \frac{1 + \cos(2\theta)}{2}$.
Show Full Solution
Square Euler's cosine definition: [M1]
$$\cos^2\theta = \left( \frac{e^{i\theta} + e^{-i\theta}}{2} \right)^2 = \frac{e^{2i\theta} + 2e^0 + e^{-2i\theta}}{4} \quad \text{[A1]}$$Group conjugate exponential terms: [M1]
$$\cos^2\theta = \frac{(e^{i(2\theta)} + e^{-i(2\theta)}) + 2}{4} = \frac{2\cos(2\theta) + 2}{4} = \frac{1 + \cos(2\theta)}{2} \quad \text{[AG]}$$Points $A, B, C$ in the Argand plane represent complex numbers $z_1, z_2, z_3$ respectively. Given that $\frac{z_3 - z_1}{z_2 - z_1} = e^{i\frac{\pi}{3}}$, prove that $\triangle ABC$ is an equilateral triangle.
Show Full Solution
1. Examine the modulus: [M1]
$$\left|\frac{z_3 - z_1}{z_2 - z_1}\right| = |e^{i\frac{\pi}{3}}| = 1 \implies \frac{|z_3 - z_1|}{|z_2 - z_1|} = 1 \implies |z_3 - z_1| = |z_2 - z_1| \quad \text{[A1]}$$This proves side lengths $AC = AB$ (isosceles triangle). [R1]
2. Examine the argument: [M1]
$$\arg\left(\frac{z_3 - z_1}{z_2 - z_1}\right) = \arg(z_3 - z_1) - \arg(z_2 - z_1) = \frac{\pi}{3} = 60^\circ \quad \text{[A1]}$$3. Geometric deduction: In $\triangle ABC$, $AB = AC$ and the included angle $\angle BAC = 60^\circ$.
The remaining two angles must be equal: $\frac{180^\circ - 60^\circ}{2} = 60^\circ$.
Therefore, all three interior angles are $60^\circ$, so $\triangle ABC$ is equilateral. [AG]
Formula Booklet References
Complex Number Forms
$$z = a + bi = r(\cos\theta + i\sin\theta) = r\text{ cis}\theta = r e^{i\theta}$$ $$|z| = r = \sqrt{a^2 + b^2}, \quad \arg(z) = \theta \in (-\pi, \pi]$$Product & Quotient Laws in Polar/Euler Form
$$|z_1 z_2| = |z_1||z_2| \quad \text{and} \quad \arg(z_1 z_2) = \arg(z_1) + \arg(z_2) \pmod{2\pi}$$ $$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|} \quad \text{and} \quad \arg\left(\frac{z_1}{z_2}\right) = \arg(z_1) - \arg(z_2) \pmod{2\pi}$$Key Formulas & Identities
1. Complex Conjugate Identities
• $z + z^* = 2\text{Re}(z), \quad z - z^* = 2i\text{Im}(z)$
• $z z^* = |z|^2 = a^2 + b^2$
• $(z_1 \pm z_2)^* = z_1^* \pm z_2^*, \quad (z_1 z_2)^* = z_1^* z_2^*, \quad \left(\frac{z_1}{z_2}\right)^* = \frac{z_1^*}{z_2^*}$
2. Euler's Trigonometric Inversions
$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$3. Geometric Loci Recognition
• $|z - z_0| = R \implies$ Circle with center $z_0$, radius $R$
• $|z - z_1| = |z - z_2| \implies$ Perpendicular bisector of segment $z_1 z_2$
• $\arg(z - z_0) = \alpha \implies$ Ray from $z_0$ at angle $\alpha$
Common Exam Pitfalls
Principal Argument Out-of-Range
Writing an argument as $\frac{4\pi}{3}$ or $240^\circ$ loses accuracy marks in IB papers. The principal range is strictly $(-\pi, \pi]$ (or $(-180^\circ, 180^\circ]$). You must subtract $2\pi$ to get $-\frac{2\pi}{3}$.
Writing $\text{Im}(z)$ as $bi$
The imaginary part $\text{Im}(z)$ is the real coefficient $b$, NOT $bi$. Stating $\text{Im}(3 + 5i) = 5i$ will be penalized by examiners.
The Quadrant Trap with $\arctan(b/a)$
Calculators only return angles in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. For $z = -1 - i$, $\frac{-1}{-1} = 1$, giving $\arctan(1) = \frac{\pi}{4}$. However, $(-1, -1)$ lies in Quadrant 3, so $\theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$. Always sketch the point!
Question 1
What is the principal argument of $z = -2 + 2i$?
Question 2
If $z = 4e^{i\frac{5\pi}{6}}$, what is its complex conjugate $z^*$ expressed in Euler's form with principal argument?
Question 3
What is the value of $(1 + i)^4$?
Question 4
In the Argand plane, the equation $|z - 3i| = 2$ represents which geometric shape?
Section B: Examination-Style Free Response Questions
Let $z = \frac{1 + i\sqrt{3}}{1 - i}$.
(a) Find the modulus and principal argument of $1 + i\sqrt{3}$ and $1 - i$. [3 marks]
(b) Hence express $z$ in polar form $r\text{ cis}\theta$. [2 marks]
(c) Find the smallest positive integer $n$ such that $z^n$ is a real number. [1 mark]
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(a) Modulus & Argument: [3 marks]
• For $1 + i\sqrt{3}$: $r_1 = \sqrt{1 + 3} = 2$ [A1], $\theta_1 = \arctan(\sqrt{3}) = \frac{\pi}{3}$ [A1].
• For $1 - i$: $r_2 = \sqrt{1 + 1} = \sqrt{2}$, $\theta_2 = -\frac{\pi}{4}$ [A1].
(b) Polar Quotient: [2 marks]
$r = \frac{r_1}{r_2} = \frac{2}{\sqrt{2}} = \sqrt{2}$ [A1].
$\theta = \theta_1 - \theta_2 = \frac{\pi}{3} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$ [A1].
$z = \sqrt{2}\text{ cis}\left(\frac{7\pi}{12}\right)$.
(c) Real condition: [1 mark]
$z^n$ is real when $\arg(z^n) = n\left(\frac{7\pi}{12}\right) = k\pi$ for $k \in \mathbb{Z}$.
$\frac{7n}{12} = k \implies 7n = 12k$. Since $7$ and $12$ are coprime, smallest positive $n = 12$. [A1]
A complex number $w$ satisfies the equation $w^3 = 8i$.
(a) Express $8i$ in polar form $r e^{i\theta}$, where $-\pi < \theta \le \pi$. [1 mark]
(b) Find the three distinct roots of $w^3 = 8i$, giving your answers in Euler's form $r e^{i\theta}$. [4 marks]
(c) Plot the three roots on an Argand diagram sketch and describe their geometric configuration. [2 marks]
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(a) Polar form of $8i$: [1 mark]
$8i = 8e^{i\frac{\pi}{2}}$ [A1].
(b) Finding the 3 roots: [4 marks]
Let $w = R e^{i\phi}$. Then $w^3 = R^3 e^{i3\phi} = 8e^{i(\frac{\pi}{2} + 2k\pi)}$ for $k = 0, 1, 2$ [M1].
$R = 8^{1/3} = 2$ [A1].
$\phi = \frac{\frac{\pi}{2} + 2k\pi}{3} = \frac{\pi}{6} + \frac{2k\pi}{3}$ [M1]:
• $k = 0$: $w_1 = 2e^{i\frac{\pi}{6}}$ [A1]
• $k = 1$: $w_2 = 2e^{i(\frac{\pi}{6} + \frac{4\pi}{6})} = 2e^{i\frac{5\pi}{6}}$ [A1]
• $k = 2$: $\frac{\pi}{6} + \frac{4\pi}{3} = \frac{9\pi}{6} = \frac{3\pi}{2}$. Normalize to $(-\pi, \pi]$: $\frac{3\pi}{2} - 2\pi = -\frac{\pi}{2}$.
$w_3 = 2e^{-i\frac{\pi}{2}}$ [A1]
(c) Geometric configuration: [2 marks]
The three roots lie on a circle centered at the origin with radius $R = 2$ [A1].
They are spaced by angular increments of $\frac{2\pi}{3} = 120^\circ$, forming the vertices of an equilateral triangle [R1].