Figure 1.3: The 6th Roots of Unity ($z^6 = 1$) Inscribed in the Unit Circle

Re Im z_0 = 1 = e^{0} z_1 = e^{i\pi/3} z_2 = e^{i 2\pi/3} z_3 = -1 = e^{i\pi} z_4 = e^{-i 2\pi/3} z_5 = e^{-i\pi/3} 2π/6 = 60°

The six roots of $z^6 = 1$ are equally spaced by $\Delta\theta = \frac{2\pi}{6} = \frac{\pi}{3}$ radians around the unit circle, forming a regular hexagon. Their vector sum is identically zero: $\sum_{k=0}^5 z_k = 0$.

AHL 1.13

De Moivre’s Theorem for Integer & Rational Exponents

  • The Core Theorem (Euler & Polar Formulation): For any angle $\theta \in \mathbb{R}$ and integer $n \in \mathbb{Z}$: $$[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta)) = r^n e^{in\theta}$$ In the compact CIS notation: $[r\operatorname{cis}\theta]^n = r^n \operatorname{cis}(n\theta)$.
  • Rigorous Mathematical Induction Proof ($n \in \mathbb{Z}^+$):
    • Base Step ($n=1$): $[r\operatorname{cis}\theta]^1 = r^1 \operatorname{cis}(1\theta)$, identically true.
    • Inductive Hypothesis: Assume true for $n=k$: $[r\operatorname{cis}\theta]^k = r^k \operatorname{cis}(k\theta)$.
    • Inductive Step ($n=k+1$): $$[r\operatorname{cis}\theta]^{k+1} = [r\operatorname{cis}\theta]^k \cdot [r\operatorname{cis}\theta] = r^k[\cos(k\theta) + i\sin(k\theta)] \cdot r[\cos\theta + i\sin\theta]$$ Expanding the algebraic product: $$= r^{k+1}[(\cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta) + i(\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta)]$$ Applying the compound angle identities $\cos(A+B)$ and $\sin(A+B)$: $$= r^{k+1}[\cos((k+1)\theta) + i\sin((k+1)\theta)] = r^{k+1}\operatorname{cis}((k+1)\theta)$$
    • Conclusion: Since true for $n=1$, and true for $n=k \implies n=k+1$, by the principle of mathematical induction it holds for all $n \in \mathbb{Z}^+$.
  • Negative Integer Exponents ($n = -m, m \in \mathbb{Z}^+$): $$z^{-m} = \frac{1}{z^m} = \frac{1}{r^m e^{im\theta}} = r^{-m} e^{-im\theta} = r^{-m}[\cos(-m\theta) + i\sin(-m\theta)] = r^{-m}[\cos(m\theta) - i\sin(m\theta)]$$
  • Rational Exponents ($q = p/q \in \mathbb{Q}$): For $w = z^{1/n}$, $w^n = z$. Since polar arguments are periodic modulo $2\pi$, raising to a fractional power yields multiple distinct branches: $$[r e^{i\theta}]^{1/n} = r^{1/n} \exp\left(i\frac{\theta + 2k\pi}{n}\right), \quad k \in \{0, 1, 2, \dots, n-1\}$$
AHL 1.13

Trigonometric Identities: Multiple Angles vs. Power Reduction

  • Type 1: Expressing Multiple Angles $\cos(n\theta), \sin(n\theta)$ in Powers of $\cos\theta, \sin\theta$: Expand $(\cos\theta + i\sin\theta)^n$ binomially using Pascal's triangle coefficients $\binom{n}{r}$, then equate real and imaginary parts: $$\cos(n\theta) = \operatorname{Re}\left[(\cos\theta + i\sin\theta)^n\right] = \sum_{k=0}^{\lfloor n/2 \rfloor} \binom{n}{2k} (-1)^k \cos^{n-2k}\theta \sin^{2k}\theta$$ $$\sin(n\theta) = \operatorname{Im}\left[(\cos\theta + i\sin\theta)^n\right] = \sum_{k=0}^{\lfloor (n-1)/2 \rfloor} \binom{n}{2k+1} (-1)^k \cos^{n-2k-1}\theta \sin^{2k+1}\theta$$ Substitute $\sin^2\theta = 1 - \cos^2\theta$ (or $\cos^2\theta = 1 - \sin^2\theta$) to obtain a polynomial purely in $\cos\theta$ or $\sin\theta$.
  • Type 2: Power Reduction of $\cos^n\theta$ and $\sin^n\theta$ into Linear Multiple Angles: Let $z = e^{i\theta} = \cos\theta + i\sin\theta$. Then $z^{-1} = e^{-i\theta} = \cos\theta - i\sin\theta$. Fundamental symmetrical pairings: $$z^k + \frac{1}{z^k} = 2\cos(k\theta), \qquad z^k - \frac{1}{z^k} = 2i\sin(k\theta)$$ To linearize powers, raise these binomials to the power $n$: $$(2\cos\theta)^n = \left(z + \frac{1}{z}\right)^n = \sum_{r=0}^n \binom{n}{r} z^{n-r} z^{-r} = \sum_{r=0}^n \binom{n}{r} z^{n-2r}$$ $$(2i\sin\theta)^n = \left(z - \frac{1}{z}\right)^n = \sum_{r=0}^n \binom{n}{r} (-1)^r z^{n-2r}$$ Pair terms with exponents $k$ and $-k$ symmetrically: $z^k + z^{-k} = 2\cos(k\theta)$. This eliminates powers and makes integration immediate without repetitive integration by parts.
AHL 1.13

$n$-th Roots of Complex Numbers & Regular Polygons

  • The Root Formula for $z^n = w$: Let $w = R e^{i\phi}$ with $R > 0$ and $\phi \in (-\pi, \pi]$. Writing $w = R e^{i(\phi + 2k\pi)}$ for $k \in \mathbb{Z}$, taking the $1/n$ power gives exactly $n$ distinct roots: $$z_k = R^{1/n} \exp\left(i \frac{\phi + 2k\pi}{n}\right), \quad k = 0, 1, 2, \dots, n-1$$
  • Geometric Properties on the Argand Plane:
    • Concentric Circle: Every root lies on the circle centered at the origin with radius $r = R^{1/n} = |w|^{1/n}$.
    • Equiangular Spacing: The angular separation between any two adjacent roots is precisely $\Delta\theta = \frac{2\pi}{n}$ radians.
    • Regular Polygon: The vertices $z_0, z_1, \dots, z_{n-1}$ form a regular $n$-sided polygon inscribed in the circle $|z| = R^{1/n}$.
    • Polygon Area Formula: Divided into $n$ congruent isosceles triangles with vertex angle $\Delta\theta = \frac{2\pi}{n}$: $$\operatorname{Area} = n \cdot \left(\frac{1}{2} r^2 \sin\left(\frac{2\pi}{n}\right)\right) = \frac{n}{2} |w|^{2/n} \sin\left(\frac{2\pi}{n}\right)$$
  • Principal Argument Constraint: The IB markscheme strictly penalizes arguments outside $(-\pi, \pi]$. If $\frac{\phi + 2k\pi}{n} > \pi$, subtract $2\pi$ to map it into the principal branch.
AHL 1.13

Roots of Unity, Cyclotomic Factorization & Trigonometric Sums

  • Definition: The $n$-th roots of unity are the solutions to $z^n = 1$. Let $\omega = e^{i\frac{2\pi}{n}} = \cos\left(\frac{2\pi}{n}\right) + i\sin\left(\frac{2\pi}{n}\right)$. The $n$ distinct roots form the cyclic group generated by $\omega$: $$\{1, \omega, \omega^2, \omega^3, \dots, \omega^{n-1}\}$$
  • The Fundamental Sum Identity: $$\sum_{k=0}^{n-1} \omega^k = 1 + \omega + \omega^2 + \dots + \omega^{n-1} = \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0 \quad (\text{since } \omega \neq 1)$$ Vector interpretation: The center of mass of the regular $n$-gon is at the origin, so the sum of vectors $\sum \vec{z_k} = \vec{0}$.
  • Factorization of $z^n - 1$: $$z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\cdots(z - \omega^{n-1}) = (z - 1)(z^{n-1} + z^{n-2} + \dots + z + 1)$$ Equating the quotient for $z \neq 1$: $$z^{n-1} + z^{n-2} + \dots + z + 1 = \prod_{k=1}^{n-1} (z - \omega^k)$$ Substituting $z = 1$ yields the famous chord product formula: $$\prod_{k=1}^{n-1} (1 - \omega^k) = n \implies \prod_{k=1}^{n-1} |1 - \omega^k| = n$$
  • Real Quadratic Factorization (Conjugate Pairing): Since $\omega^{n-k} = \overline{\omega^k}$, roots pair as complex conjugate pairs: $$(z - \omega^k)(z - \overline{\omega^k}) = z^2 - 2\operatorname{Re}(\omega^k) z + |\omega^k|^2 = z^2 - 2z\cos\left(\frac{2k\pi}{n}\right) + 1$$ This factors $z^n - 1$ into real linear and quadratic factors.

Authentic IB HL Worked Exemplars (Full Markscheme Rubrics)

Exemplar 1 · Paper 1 (6 Marks)

Rational Complex Fraction to a High Exponent

Consider the complex numbers $z_1 = 1 + i\sqrt{3}$ and $z_2 = 1 - i$.
(a) Express $z_1$ and $z_2$ in modulus-argument form. [2 marks]
(b) Hence evaluate $\frac{z_1^8}{z_2^6}$, expressing your answer in the Cartesian form $a + bi$, where $a, b \in \mathbb{R}$. [4 marks]

Full Markscheme & Method:

  1. Part (a): Modulus-Argument conversions
    For $z_1 = 1 + i\sqrt{3}$: $$|z_1| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2, \quad \operatorname{Arg}(z_1) = \arctan(\sqrt{3}) = \frac{\pi}{3} \quad (\text{Quadrant 1})$$ $$z_1 = 2 e^{i\frac{\pi}{3}} = 2\operatorname{cis}\left(\frac{\pi}{3}\right) \quad \mathbf{[A1]}$$ For $z_2 = 1 - i$: $$|z_2| = \sqrt{1^2 + (-1)^2} = \sqrt{2}, \quad \operatorname{Arg}(z_2) = \arctan\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \quad (\text{Quadrant 4})$$ $$z_2 = \sqrt{2} e^{-i\frac{\pi}{4}} = \sqrt{2}\operatorname{cis}\left(-\frac{\pi}{4}\right) \quad \mathbf{[A1]}$$
  2. Part (b): Applying De Moivre’s Theorem $$z_1^8 = \left(2 e^{i\frac{\pi}{3}}\right)^8 = 2^8 e^{i\frac{8\pi}{3}} = 256 e^{i\frac{2\pi}{3}} \quad \mathbf{[M1]}$$ $$z_2^6 = \left(\sqrt{2} e^{-i\frac{\pi}{4}}\right)^6 = (\sqrt{2})^6 e^{-i\frac{6\pi}{4}} = 8 e^{-i\frac{3\pi}{2}} = 8 e^{i\frac{\pi}{2}} \quad \mathbf{[M1]}$$
  3. Dividing the terms: $$W = \frac{z_1^8}{z_2^6} = \frac{256 e^{i\frac{2\pi}{3}}}{8 e^{i\frac{\pi}{2}}} = \frac{256}{8} e^{i\left(\frac{2\pi}{3} - \frac{\pi}{2}\right)} = 32 e^{i\frac{\pi}{6}} \quad \mathbf{[A1]}$$
  4. Convert to Cartesian form $a + bi$: $$W = 32\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right) = 32\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = 16\sqrt{3} + 16i \quad \mathbf{[A1]}$$
Examiner Advice: Never attempt binomial expansion directly on $(1+i\sqrt{3})^8$ in Paper 1. Conversion to polar/Euler form is the only acceptable route according to the markscheme `[M1]`. Always check whether the final answer is requested in Cartesian ($a+bi$) or Euler ($re^{i\theta}$) form.
Exemplar 2 · Paper 1 (7 Marks)

Multiple-Angle Identity & Quartic Polynomial Solving

(a) By expanding $(\cos\theta + i\sin\theta)^4$ using the Binomial Theorem and De Moivre's Theorem, prove that: $$\cos(4\theta) = 8\cos^4\theta - 8\cos^2\theta + 1 \span class="mark-allocation">[4 marks]$$ (b) Hence solve the equation $8x^4 - 8x^2 + 1 = 0$, giving your answers in exact trigonometric form. [3 marks]

Full Markscheme & Method:

  1. Part (a): Binomial & De Moivre Expansion By De Moivre’s Theorem: $$\cos(4\theta) + i\sin(4\theta) = (\cos\theta + i\sin\theta)^4 \quad \mathbf{[M1]}$$ Expanding binomially with coefficients $1, 4, 6, 4, 1$: $$= \cos^4\theta + 4\cos^3\theta(i\sin\theta) + 6\cos^2\theta(i\sin\theta)^2 + 4\cos\theta(i\sin\theta)^3 + (i\sin\theta)^4$$ Using $i^2 = -1, i^3 = -i, i^4 = 1$: $$= (\cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta) + i(4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta) \quad \mathbf{[A1]}$$
  2. Equating Real Parts: $$\cos(4\theta) = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta \quad \mathbf{[M1]}$$ Substitute $\sin^2\theta = 1 - \cos^2\theta$ and $\sin^4\theta = (1 - \cos^2\theta)^2 = 1 - 2\cos^2\theta + \cos^4\theta$: $$\cos(4\theta) = \cos^4\theta - 6\cos^2\theta(1 - \cos^2\theta) + (1 - 2\cos^2\theta + \cos^4\theta)$$ $$= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + 1 - 2\cos^2\theta + \cos^4\theta$$ $$= 8\cos^4\theta - 8\cos^2\theta + 1 \quad \mathbf{[AG][R1]}$$
  3. Part (b): Solving $8x^4 - 8x^2 + 1 = 0$ Let $x = \cos\theta$. Then the equation becomes: $$\cos(4\theta) = 0 \quad \mathbf{[M1]}$$ Solve for $4\theta \in [0, 4\pi)$ to yield 4 distinct values: $$4\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2} \implies \theta = \frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8} \quad \mathbf{[A1]}$$ Since $\cos(5\pi/8) = -\cos(3\pi/8)$ and $\cos(7\pi/8) = -\cos(\pi/8)$, the four exact solutions are: $$x = \pm \cos\left(\frac{\pi}{8}\right), \quad x = \pm \cos\left(\frac{3\pi}{8}\right) \quad \mathbf{[A1]}$$
Markscheme Trap Alert: The command term "Hence" means you MUST link the polynomial $8x^4 - 8x^2 + 1 = 0$ to the identity derived in part (a). Attempting the quadratic formula on $u = x^2$ scores ZERO for part (b) in IB exams.
Exemplar 3 · Paper 1 (7 Marks)

Power Reduction via $z - z^{-1}$ & Definite Integration

(a) Given $z = e^{i\theta}$, express $(2i\sin\theta)^5$ in terms of $z$ and $z^{-1}$. Hence show that: $$\sin^5\theta = \frac{1}{16}(\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta) \span class="mark-allocation">[4 marks]$$ (b) Hence evaluate the definite integral $\int_0^{\pi/2} \sin^5\theta \, d\theta$. [3 marks]

Full Markscheme & Method:

  1. Part (a): Symmetrical Binomial Expansion Using $z - z^{-1} = 2i\sin\theta$: $$(2i\sin\theta)^5 = \left(z - \frac{1}{z}\right)^5 \quad \mathbf{[M1]}$$ Expand using binomial coefficients $\binom{5}{0}=1, \binom{5}{1}=5, \binom{5}{2}=10, \binom{5}{3}=10, \binom{5}{4}=5, \binom{5}{5}=1$: $$= z^5 - 5z^3 + 10z - 10z^{-1} + 5z^{-3} - z^{-5} \quad \mathbf{[A1]}$$ Group symmetric terms: $$= (z^5 - z^{-5}) - 5(z^3 - z^{-3}) + 10(z - z^{-1})$$ Substitute $z^k - z^{-k} = 2i\sin(k\theta)$: $$= 2i\sin(5\theta) - 5(2i\sin(3\theta)) + 10(2i\sin\theta) = 2i[\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta] \quad \mathbf{[M1]}$$
  2. Simplify LHS: $$(2i\sin\theta)^5 = 2^5 i^5 \sin^5\theta = 32(i) \sin^5\theta = 32i \sin^5\theta$$ Equating LHS and RHS: $$32i\sin^5\theta = 2i[\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta]$$ Divide both sides by $32i$: $$\sin^5\theta = \frac{1}{16}(\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta) \quad \mathbf{[AG][R1]}$$
  3. Part (b): Evaluating the Definite Integral $$\int_0^{\pi/2} \sin^5\theta \, d\theta = \frac{1}{16} \int_0^{\pi/2} (\sin(5\theta) - 5\sin(3\theta) + 10\sin\theta) \, d\theta \quad \mathbf{[M1]}$$ Integrating each trigonometric term: $$= \frac{1}{16} \left[ -\frac{1}{5}\cos(5\theta) + \frac{5}{3}\cos(3\theta) - 10\cos\theta \right]_0^{\pi/2} \quad \mathbf{[A1]}$$ Upper limit $\theta = \frac{\pi}{2}$: $\cos\left(\frac{5\pi}{2}\right) = 0, \quad \cos\left(\frac{3\pi}{2}\right) = 0, \quad \cos\left(\frac{\pi}{2}\right) = 0 \implies \text{Upper} = 0$.
    Lower limit $\theta = 0$: $\cos(0) = 1 \implies \text{Lower} = -\frac{1}{5} + \frac{5}{3} - 10 = \frac{-3 + 25 - 150}{15} = -\frac{128}{15}$.
    Subtracting limits: $$\text{Value} = \frac{1}{16}\left(0 - \left(-\frac{128}{15}\right)\right) = \frac{1}{16} \times \frac{128}{15} = \frac{8}{15} \quad \mathbf{[A1]}$$
Key Checking Technique: In Paper 2, you can verify this immediately with GDC: $\int_0^{\pi/2} \sin^5(x)dx \approx 0.533333 = \frac{8}{15}$. In Paper 1, double check arithmetic on the lower limit fraction!
Exemplar 4 · Paper 2 (8 Marks)

Roots of $z^5 = -16\sqrt{3} + 16i$ & Polygon Verification

Consider the equation $z^5 = -16\sqrt{3} + 16i$.
(a) Express $-16\sqrt{3} + 16i$ in the form $r e^{i\theta}$, where $r > 0$ and $-\pi < \theta \le \pi$. [2 marks]
(b) Find all five solutions to the equation, giving your answers in the form $R e^{i\alpha}$ where $-\pi < \alpha \le \pi$. [4 marks]
(c) Show that the sum of all five roots is zero. [2 marks]

Full Markscheme & Method:

  1. Part (a): Polar Form of RHS $$r = \sqrt{(-16\sqrt{3})^2 + 16^2} = \sqrt{256 \times 3 + 256} = \sqrt{256 \times 4} = 16 \times 2 = 32 \quad \mathbf{[A1]}$$ Since $x < 0$ and $y > 0$, the point lies in Quadrant 2: $$\theta = \pi - \arctan\left(\frac{16}{16\sqrt{3}}\right) = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \quad \mathbf{[A1]}$$ $$-16\sqrt{3} + 16i = 32 e^{i\frac{5\pi}{6}}$$
  2. Part (b): Solving $z^5 = 32 e^{i(5\pi/6 + 2k\pi)}$ Take the $1/5$ power: $$z_k = 32^{1/5} \exp\left(i \frac{5\pi/6 + 2k\pi}{5}\right) = 2 \exp\left(i\left(\frac{\pi}{6} + \frac{2k\pi}{5}\right)\right) \quad \mathbf{[M1]}$$ Angular step: $\frac{2\pi}{5} = \frac{12\pi}{30}$; starting angle $\frac{\pi}{6} = \frac{5\pi}{30}$. Evaluate for $k = 0, 1, 2, -1, -2$ (or $k=0,1,2,3,4$ mapped to $(-\pi, \pi]$):
    • $k=0$: $\alpha_0 = \frac{\pi}{6}$ $\implies z_0 = 2e^{i\frac{\pi}{6}}$ $\mathbf{[A1]}$
    • $k=1$: $\alpha_1 = \frac{5\pi}{30} + \frac{12\pi}{30} = \frac{17\pi}{30}$ $\implies z_1 = 2e^{i\frac{17\pi}{30}}$ $\mathbf{[A1]}$
    • $k=2$: $\alpha_2 = \frac{17\pi}{30} + \frac{12\pi}{30} = \frac{29\pi}{30}$ $\implies z_2 = 2e^{i\frac{29\pi}{30}}$ $\mathbf{[A1]}$
    • $k=-1$: $\alpha_3 = \frac{5\pi}{30} - \frac{12\pi}{30} = -\frac{7\pi}{30}$ $\implies z_3 = 2e^{-i\frac{7\pi}{30}}$
    • $k=-2$: $\alpha_4 = -\frac{7\pi}{30} - \frac{12\pi}{30} = -\frac{19\pi}{30}$ $\implies z_4 = 2e^{-i\frac{19\pi}{30}}$ $\mathbf{[A1]}$
  3. Part (c): Proving the Sum of Roots is Zero Method 1 (Polynomial / Vieta's Theorem): The roots of $z^5 - (-16\sqrt{3} + 16i) = 0$ are the roots of a polynomial of the form $z^5 + 0z^4 + 0z^3 + 0z^2 + 0z - w = 0$. By Vieta's formulas, the sum of roots is given by $-\frac{a_{n-1}}{a_n} = -\frac{0}{1} = 0 \quad \mathbf{[M1][R1]}$.
    Method 2 (Geometric Series): Notice $z_k = z_0 \cdot \omega^k$ where $\omega = e^{i\frac{2\pi}{5}}$. $$\sum_{k=0}^4 z_k = z_0 \sum_{k=0}^4 \omega^k = z_0 \left(\frac{1 - \omega^5}{1 - \omega}\right) = z_0 \left(\frac{1 - 1}{1 - \omega}\right) = 0 \quad \mathbf{[M1][AG]}$$
Exemplar 5 · Paper 3 (9 Marks)

Roots of Unity, Cyclotomic Reductions & Exact Cosine Sums

Let $\omega = e^{i\frac{2\pi}{7}}$.
(a) Show that $1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0$. [2 marks]
(b) By pairing conjugate powers of $\omega$, prove that: $$\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) = -\frac{1}{2} \span class="mark-allocation">[4 marks]$$ (c) Hence deduce that: $$\cos\left(\frac{\pi}{7}\right) - \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{3\pi}{7}\right) = \frac{1}{2} \span class="mark-allocation">[3 marks]$$

Full Markscheme & Method:

  1. Part (a): Verification of Sum of 7th Roots of Unity Since $\omega = e^{i\frac{2\pi}{7}}$, $\omega^7 = e^{i 2\pi} = 1$. $$\omega^7 - 1 = 0 \implies (\omega - 1)(1 + \omega + \omega^2 + \dots + \omega^6) = 0 \quad \mathbf{[M1]}$$ Since $\omega \neq 1$, we divide by $(\omega - 1)$: $$1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0 \quad \mathbf{[AG][R1]}$$
  2. Part (b): Pairing Conjugate Roots Rearrange the sum: $$(\omega + \omega^6) + (\omega^2 + \omega^5) + (\omega^3 + \omega^4) = -1 \quad \mathbf{[M1]}$$ Notice that $\omega^6 = \omega^{-1} = \overline{\omega}$, $\omega^5 = \omega^{-2} = \overline{\omega^2}$, and $\omega^4 = \omega^{-3} = \overline{\omega^3} \quad \mathbf{[R1]}$. Using the identity $z + \overline{z} = 2\operatorname{Re}(z) = 2\cos(\operatorname{Arg} z)$: $$\omega + \omega^6 = 2\cos\left(\frac{2\pi}{7}\right)$$ $$\omega^2 + \omega^5 = 2\cos\left(\frac{4\pi}{7}\right)$$ $$\omega^3 + \omega^4 = 2\cos\left(\frac{6\pi}{7}\right) \quad \mathbf{[A1]}$$ Summing these three pairs: $$2\left[\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)\right] = -1$$ Dividing by 2 gives: $$\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) = -\frac{1}{2} \quad \mathbf{[AG][A1]}$$
  3. Part (c): Supplementary Angle Transformation Recall the supplementary angle identity $\cos(\pi - \theta) = -\cos\theta$: $$\cos\left(\frac{6\pi}{7}\right) = \cos\left(\pi - \frac{\pi}{7}\right) = -\cos\left(\frac{\pi}{7}\right) \quad \mathbf{[M1]}$$ $$\cos\left(\frac{4\pi}{7}\right) = \cos\left(\pi - \frac{3\pi}{7}\right) = -\cos\left(\frac{3\pi}{7}\right) \quad \mathbf{[M1]}$$ Substitute these into the relation from part (b): $$\cos\left(\frac{2\pi}{7}\right) - \cos\left(\frac{3\pi}{7}\right) - \cos\left(\frac{\pi}{7}\right) = -\frac{1}{2}$$ Multiplying both sides by $-1$: $$\cos\left(\frac{\pi}{7}\right) - \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{3\pi}{7}\right) = \frac{1}{2} \quad \mathbf{[AG][A1]}$$
Tier 1: Powers & Basic Roots (Questions 1–4)
Exercise 1 Paper 1 · 4 Marks

Using De Moivre’s theorem, express $(\sqrt{3} + i)^6$ in the form $a + bi$, where $a, b \in \mathbb{R}$.

View Complete Solution & Markscheme
  1. Find modulus and argument of $z = \sqrt{3} + i$: $$r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \quad \mathbf{[A1]}$$ Since $x > 0, y > 0$ (Quadrant 1): $$\theta = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \implies z = 2\operatorname{cis}\left(\frac{\pi}{6}\right) = 2e^{i\frac{\pi}{6}} \quad \mathbf{[A1]}$$
  2. Apply De Moivre’s theorem with $n=6$: $$z^6 = 2^6\operatorname{cis}\left(6 \times \frac{\pi}{6}\right) = 64\operatorname{cis}(\pi) = 64 e^{i\pi} \quad \mathbf{[M1]}$$
  3. Evaluate $\operatorname{cis}(\pi)$ and state in $a+bi$ form: $$\operatorname{cis}(\pi) = \cos(\pi) + i\sin(\pi) = -1 + 0i$$ $$z^6 = 64(-1 + 0i) = -64 + 0i \quad (\text{or } -64) \quad \mathbf{[A1]}$$

Final Answer: $-64$ (or $-64 + 0i$)

Exercise 2 Paper 1 · 3 Marks

Evaluate $\left(\cos\left(\frac{\pi}{5}\right) + i\sin\left(\frac{\pi}{5}\right)\right)^{10}$ without the use of a calculator.

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  1. Apply De Moivre’s theorem directly: $$\left(\cos\left(\frac{\pi}{5}\right) + i\sin\left(\frac{\pi}{5}\right)\right)^{10} = \cos\left(10 \times \frac{\pi}{5}\right) + i\sin\left(10 \times \frac{\pi}{5}\right) \quad \mathbf{[M1]}$$
  2. Simplify the angle: $$10 \times \frac{\pi}{5} = 2\pi \quad \mathbf{[A1]}$$
  3. Evaluate exact values: $$\cos(2\pi) + i\sin(2\pi) = 1 + 0i = 1 \quad \mathbf{[A1]}$$

Final Answer: $1$

Exercise 3 Paper 1 · 5 Marks

Find all four solutions to the equation $z^4 = -16$, expressing each root in the form $r e^{i\theta}$ where $-\pi < \theta \le \pi$.

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  1. Write the right-hand side $-16$ in general polar/exponential form: $$-16 = 16 e^{i(\pi + 2k\pi)}, \quad k \in \mathbb{Z} \quad \mathbf{[M1]}$$
  2. Extract the 4th root: $$|z| = 16^{1/4} = 2 \quad \mathbf{[A1]}$$ $$\operatorname{Arg}(z_k) = \frac{\pi + 2k\pi}{4} = \frac{\pi}{4} + \frac{k\pi}{2} \quad \mathbf{[M1]}$$
  3. Evaluate for $k = 0, 1, 2, 3$ (or $k=0,1,-1,-2$):
    • $k = 0$: $z_0 = 2e^{i\frac{\pi}{4}}$
    • $k = 1$: $z_1 = 2e^{i\frac{3\pi}{4}}$
    • $k = 2$: $\frac{5\pi}{4} \equiv -\frac{3\pi}{4} \implies z_2 = 2e^{-i\frac{3\pi}{4}}$
    • $k = 3$: $\frac{7\pi}{4} \equiv -\frac{\pi}{4} \implies z_3 = 2e^{-i\frac{\pi}{4}} \quad \mathbf{[A2]}$

Final Answer: $2e^{i\frac{\pi}{4}}, 2e^{i\frac{3\pi}{4}}, 2e^{-i\frac{3\pi}{4}}, 2e^{-i\frac{\pi}{4}}$

Exercise 4 Paper 1 · 4 Marks

Let $\omega$ be a non-real root of $z^3 = 1$. Simplify the expression $(2 + 5\omega + 2\omega^2)^3$.

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  1. Recall that $\omega$ satisfies $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1 \quad \mathbf{[R1]}$.
  2. Rearrange the inner expression: $$2 + 5\omega + 2\omega^2 = 2(1 + \omega + \omega^2) + 3\omega \quad \mathbf{[M1]}$$ $$= 2(0) + 3\omega = 3\omega \quad \mathbf{[A1]}$$
  3. Cube the simplified expression: $$(3\omega)^3 = 3^3 \cdot \omega^3 = 27 \cdot (1) = 27 \quad \mathbf{[A1]}$$

Final Answer: $27$

Tier 2: Trigonometric Identities & Root Geometry (Questions 5–8)
Exercise 5 Paper 1 · 5 Marks

Using De Moivre’s theorem and the Binomial Theorem, prove that $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$.

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  1. Apply De Moivre’s theorem: $$\cos(3\theta) + i\sin(3\theta) = (\cos\theta + i\sin\theta)^3 \quad \mathbf{[M1]}$$
  2. Expand the RHS binomially: $$= \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3$$ $$= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta) \quad \mathbf{[A1]}$$
  3. Equate imaginary parts: $$\sin(3\theta) = 3\cos^2\theta\sin\theta - \sin^3\theta \quad \mathbf{[M1]}$$
  4. Substitute $\cos^2\theta = 1 - \sin^2\theta$: $$\sin(3\theta) = 3(1 - \sin^2\theta)\sin\theta - \sin^3\theta = 3\sin\theta - 3\sin^3\theta - \sin^3\theta$$ $$= 3\sin\theta - 4\sin^3\theta \quad \mathbf{[AG][R1]}$$

Final Answer: Identity proven algebraically.

Exercise 6 Paper 1 · 6 Marks

Show that $\cos^4\theta = \frac{1}{8}(\cos(4\theta) + 4\cos(2\theta) + 3)$, and hence find the exact value of $\int_0^{\pi/2} \cos^4\theta\,d\theta$.

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  1. Let $z = e^{i\theta}$, so $2\cos\theta = z + z^{-1}$: $$(2\cos\theta)^4 = 16\cos^4\theta = \left(z + \frac{1}{z}\right)^4 \quad \mathbf{[M1]}$$ $$= z^4 + 4z^2 + 6 + 4z^{-2} + z^{-4} = (z^4 + z^{-4}) + 4(z^2 + z^{-2}) + 6 \quad \mathbf{[A1]}$$
  2. Substitute $z^k + z^{-k} = 2\cos(k\theta)$: $$16\cos^4\theta = 2\cos(4\theta) + 4(2\cos(2\theta)) + 6 = 2\cos(4\theta) + 8\cos(2\theta) + 6 \quad \mathbf{[M1]}$$ Divide by 16: $$\cos^4\theta = \frac{1}{8}(\cos(4\theta) + 4\cos(2\theta) + 3) \quad \mathbf{[AG]}$$
  3. Integrate term by term: $$\int_0^{\pi/2} \cos^4\theta\,d\theta = \frac{1}{8}\left[ \frac{1}{4}\sin(4\theta) + 2\sin(2\theta) + 3\theta \right]_0^{\pi/2} \quad \mathbf{[M1]}$$ At $\theta = \pi/2$: $\sin(2\pi) = 0, \sin(\pi) = 0$, giving $\frac{1}{8}\left(0 + 0 + \frac{3\pi}{2}\right) = \frac{3\pi}{16} \quad \mathbf{[A1]}$

Final Answer: $\frac{3\pi}{16}$

Exercise 7 Paper 1 · 5 Marks

The roots of the equation $z^8 = 256$ form the vertices of a regular octagon in the Argand plane.
Find the exact area of this octagon.

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  1. Find the circumradius $R$: $$R = |z| = 256^{1/8} = (2^8)^{1/8} = 2 \quad \mathbf{[A1]}$$
  2. Central angle between consecutive vertices: $$\Delta\theta = \frac{2\pi}{8} = \frac{\pi}{4} \quad \mathbf{[A1]}$$
  3. Area of one triangle with two sides of length $R$ and included angle $\Delta\theta$: $$A_1 = \frac{1}{2} R^2 \sin\left(\frac{\pi}{4}\right) = \frac{1}{2}(2^2)\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2} \quad \mathbf{[M1]}$$
  4. Total octagon area (8 identical triangles): $$\text{Area} = 8 \times A_1 = 8\sqrt{2} \quad \mathbf{[A1]}$$

Final Answer: $8\sqrt{2}$

Exercise 8 Paper 1 · 6 Marks

Given that $\omega = e^{i\frac{2\pi}{5}}$ is a fifth root of unity, show that: $$1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$$ Hence prove that $\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2}$.

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  1. $\omega^5 = 1 \implies \omega^5 - 1 = 0 \implies (\omega - 1)(1 + \omega + \omega^2 + \omega^3 + \omega^4) = 0$. Since $\omega \neq 1$, $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0 \quad \mathbf{[M1][AG]}$.
  2. Take real parts of the summation: $$\operatorname{Re}(1 + \omega + \omega^2 + \omega^3 + \omega^4) = 1 + \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) + \cos\left(\frac{6\pi}{5}\right) + \cos\left(\frac{8\pi}{5}\right) = 0 \quad \mathbf{[M1]}$$
  3. Use cosine symmetry $\cos(2\pi - \theta) = \cos\theta$: $$\cos\left(\frac{8\pi}{5}\right) = \cos\left(\frac{2\pi}{5}\right), \quad \cos\left(\frac{6\pi}{5}\right) = \cos\left(\frac{4\pi}{5}\right) \quad \mathbf{[R1]}$$ $$1 + 2\cos\left(\frac{2\pi}{5}\right) + 2\cos\left(\frac{4\pi}{5}\right) = 0 \quad \mathbf{[A1]}$$
  4. Rearrange for the required sum: $$2\left[\cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right)\right] = -1 \implies \cos\left(\frac{2\pi}{5}\right) + \cos\left(\frac{4\pi}{5}\right) = -\frac{1}{2} \quad \mathbf{[AG][A1]}$$

Final Answer: Proven: $-\frac{1}{2}$

Tier 3: Paper 3 Style Investigations (Questions 9–10)
Exercise 9 Paper 3 Extended Investigation · 7 Marks

Let $\omega_k = e^{i\frac{2k\pi}{n}}$ for $k = 0, 1, \dots, n-1$ be the $n$th roots of unity.
(a) Explain why $z^n - 1 = (z - 1)(z - \omega)(z - \omega^2)\dots(z - \omega^{n-1})$. [2 marks]
(b) By dividing both sides by $(z - 1)$ and taking the limit as $z \to 1$, evaluate the product: $$(1 - \omega)(1 - \omega^2)(1 - \omega^3)\dots(1 - \omega^{n-1}) \span class="mark-allocation">[5 marks]$$

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  1. Part (a): By the Fundamental Theorem of Algebra, any monic polynomial of degree $n$ factors as $\prod_{k=0}^{n-1}(z - r_k)$. Since $\omega^k$ are the $n$ distinct roots of $z^n - 1 = 0$, the identity follows directly $\mathbf{[R1][AG]}$.
  2. Part (b): Division and Geometric series $$\frac{z^n - 1}{z - 1} = z^{n-1} + z^{n-2} + \dots + z + 1 = \prod_{k=1}^{n-1} (z - \omega^k) \quad \mathbf{[M1]}$$
  3. Take the limit as $z \to 1$ (or substitute $z = 1$ by polynomial continuity): $$\lim_{z \to 1} (z^{n-1} + z^{n-2} + \dots + z + 1) = \underbrace{1 + 1 + \dots + 1}_{n \text{ terms}} = n \quad \mathbf{[M1][A1]}$$
  4. Equating with the product expression: $$\prod_{k=1}^{n-1} (1 - \omega^k) = n \quad \mathbf{[AG][A1]}$$

Final Answer: $(1 - \omega)(1 - \omega^2)\dots(1 - \omega^{n-1}) = n$

Exercise 10 Paper 1 / Paper 3 · 8 Marks

Prove that for $\theta \neq 2k\pi$: $$\sum_{r=0}^n \cos(r\theta) = \frac{\sin\left(\frac{(n+1)\theta}{2}\right)\cos\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}$$

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  1. Consider the complex geometric series: $$S = \sum_{r=0}^n e^{ir\theta} = 1 + e^{i\theta} + e^{i2\theta} + \dots + e^{in\theta} \quad \mathbf{[M1]}$$ Using the finite GP sum formula $S = \frac{1 - z^{n+1}}{1 - z}$ where $z = e^{i\theta}$: $$S = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}} \quad \mathbf{[A1]}$$
  2. Factor out half-angles from numerator and denominator: $$1 - e^{i\phi} = e^{i\phi/2}(e^{-i\phi/2} - e^{i\phi/2}) = e^{i\phi/2}(-2i\sin(\phi/2)) \quad \mathbf{[M1]}$$ Applying this to numerator and denominator: $$S = \frac{e^{i(n+1)\theta/2}\left(-2i\sin\left(\frac{(n+1)\theta}{2}\right)\right)}{e^{i\theta/2}\left(-2i\sin\left(\frac{\theta}{2}\right)\right)} = e^{i n\theta/2} \cdot \frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \quad \mathbf{[A1]}$$
  3. Take the real part of both sides: $$\operatorname{Re}(S) = \sum_{r=0}^n \cos(r\theta) \quad \text{and} \quad \operatorname{Re}\left(e^{in\theta/2}\right) = \cos\left(\frac{n\theta}{2}\right) \quad \mathbf{[M1]}$$ $$\sum_{r=0}^n \cos(r\theta) = \frac{\sin\left(\frac{(n+1)\theta}{2}\right)\cos\left(\frac{n\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)} \quad \mathbf{[AG][A1]}$$

Final Answer: Formula proven via complex geometric series summation.

Formula Booklet AHL 1.13

De Moivre’s Theorem

  • Integer & Rational Powers: $$[r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta)) = r^n e^{in\theta}$$
  • Roots of a Complex Number: For $z^n = w = r e^{i\theta}$: $$z_k = r^{1/n} \exp\left(i \frac{\theta + 2k\pi}{n}\right), \quad k = 0, 1, \dots, n-1$$
Must Memorize

Symmetric Groupings & Roots of Unity

  • Euler Power Pairing: $$z^k + \frac{1}{z^k} = 2\cos(k\theta), \qquad z^k - \frac{1}{z^k} = 2i\sin(k\theta)$$
  • Roots of Unity Sum & Product: $$\omega_k = e^{i\frac{2k\pi}{n}}, \quad \sum_{k=0}^{n-1} \omega^k = 0, \quad \prod_{k=0}^{n-1} \omega_k = (-1)^{n-1}$$
  • Inscribed Regular $n$-gon Area: $$\text{Area} = \frac{n}{2} |w|^{2/n} \sin\left(\frac{2\pi}{n}\right)$$

Critical IB Examination Pitfalls in De Moivre’s Theorem

  • Forgetting to Exponentiate the Modulus: Writing $[2\operatorname{cis}\theta]^4 = 2\operatorname{cis}(4\theta)$ instead of $2^4\operatorname{cis}(4\theta) = 16\operatorname{cis}(4\theta)$ immediately loses 2 marks.
  • Missing Distinct Roots (+2kπ): Forgetting $+2k\pi$ in $z^n = w$ gives only 1 of the $n$ roots. An $n$-th degree complex polynomial equation ALWAYS yields exactly $n$ distinct roots on a circle.
  • Principal Argument Constraint: Leaving root arguments outside $(-\pi, \pi]$ (e.g., leaving $\theta = \frac{11\pi}{8}$ instead of $-\frac{5\pi}{8}$) triggers the syllabus boundary penalty.
  • Powers of $i$ in Binomial Reductions: A single sign slip in $i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i$ invalidates your real/imaginary separation and costs all subsequent method marks.

Section A: Diagnostic Check

4 core concept check questions. Immediate feedback and scoring.

Score: 0 / 4

Question 1

What is the value of $\left(\cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right)\right)^3$?

Correct! $3 \times \frac{\pi}{6} = \frac{\pi}{2}$, and $\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right) = 0 + i(1) = i$.

Question 2

If $\omega$ is a primitive cube root of unity, what is the value of $1 + \omega + \omega^2$?

Correct! The sum of all $n$ roots of unity is always zero: $\sum_{k=0}^{n-1} \omega^k = 0$.

Question 3

The roots of $z^5 = 32$ form the vertices of a regular pentagon in the Argand plane. What is the radius of the circumscribed circle?

Correct! The modulus of every root is $R = 32^{1/5} = 2$.

Question 4

What is the angle between adjacent roots of $z^8 = 1$ in the Argand plane?

Correct! The angular separation is $\Delta\theta = \frac{2\pi}{n} = \frac{2\pi}{8} = \frac{\pi}{4}$.

Section B: Examination-Style Free Response Questions

Question 5 (Paper 1 Style · 6 marks)

Let $z = \cos\theta + i\sin\theta$.
(a) Show that $z^n + z^{-n} = 2\cos(n\theta)$ and $z^n - z^{-n} = 2i\sin(n\theta)$. [2 marks]
(b) Hence express $\sin^3\theta$ in the form $A\sin(3\theta) + B\sin\theta$, where $A, B \in \mathbb{Q}$. [4 marks]

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(a) Verification: [2 marks]
$z^n = \cos(n\theta) + i\sin(n\theta)$, $z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta)$ [M1].
Sum $= 2\cos(n\theta)$ [A1]; Difference $= 2i\sin(n\theta)$ [A1].

(b) Expansion of $(2i\sin\theta)^3$: [4 marks]
$(2i\sin\theta)^3 = -8i\sin^3\theta = (z - z^{-1})^3$ [M1]
$= z^3 - 3z^2(z^{-1}) + 3z(z^{-2}) - z^{-3} = (z^3 - z^{-3}) - 3(z - z^{-1})$ [M1]
$= 2i\sin(3\theta) - 3(2i\sin\theta) = 2i\sin(3\theta) - 6i\sin\theta$ [A1]
Divide by $-8i$:
$\sin^3\theta = \frac{2i\sin(3\theta) - 6i\sin\theta}{-8i} = -\frac{1}{4}\sin(3\theta) + \frac{3}{4}\sin\theta$ [A1].

Question 6 (Paper 2 Style · 8 marks)

Consider the complex equation $(z + 1)^4 = z^4$.
(a) Show that any solution $z$ must satisfy $\left|\frac{z+1}{z}\right| = 1$, and deduce that all solutions have $\text{Re}(z) = -\frac{1}{2}$. [3 marks]
(b) By writing $\frac{z+1}{z} = e^{i\theta}$, find all three solutions to the equation. [5 marks]

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(a) Modulus & Real part deduction: [3 marks]
$|(z+1)^4| = |z^4| \implies |z+1|^4 = |z|^4 \implies |z+1| = |z|$, so $\left|\frac{z+1}{z}\right| = 1$ [M1].
$|z - (-1)| = |z - 0|$ represents points equidistant from $0$ and $-1$ in the Argand plane [M1].
The perpendicular bisector of the segment from $-1$ to $0$ is the vertical line $x = -\frac{1}{2}$, so $\text{Re}(z) = -\frac{1}{2}$ [A1].

(b) Finding the 3 solutions: [5 marks]
$\left(\frac{z+1}{z}\right)^4 = 1 \implies \frac{z+1}{z} = e^{i\frac{2k\pi}{4}} = e^{i\frac{k\pi}{2}}$ for $k = 1, 2, 3$ ($k=0$ gives $\frac{z+1}{z}=1 \implies 1=0$, impossible) [M1].
$1 + \frac{1}{z} = e^{i\frac{k\pi}{2}} \implies \frac{1}{z} = e^{i\frac{k\pi}{2}} - 1 \implies z = \frac{1}{e^{i\frac{k\pi}{2}} - 1}$ [M1]:
• $k = 1$: $e^{i\pi/2} = i \implies z_1 = \frac{1}{i - 1} = \frac{-1 - i}{2} = -\frac{1}{2} - \frac{1}{2}i$ [A1]
• $k = 2$: $e^{i\pi} = -1 \implies z_2 = \frac{1}{-1 - 1} = -\frac{1}{2}$ [A1]
• $k = 3$: $e^{i3\pi/2} = -i \implies z_3 = \frac{1}{-i - 1} = \frac{-1 + i}{2} = -\frac{1}{2} + \frac{1}{2}i$ [A1].
All three solutions have $\text{Re}(z) = -\frac{1}{2}$.