# Complex Numbers & Functions
> A quantum amplitude is not a probability — it is a *complex number*. Probabilities are the
> $|\cdot|^2$ shadow it casts. Everything distinctively quantum, from interference to the Fourier
> transform that powers Shor's algorithm, lives in the *phase* that the shadow throws away. This
> lesson rebuilds the complex field from the ground up so that phase is never mysterious again.
## Learning Objectives
By the end of this lesson you will be able to:
1. **Manipulate** complex numbers fluently in both Cartesian ($a+bi$) and polar ($re^{i\theta}$) form, converting between them and computing modulus, argument, and conjugate.
2. **Derive** Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$ from the Maclaurin series and use it to prove de Moivre's theorem.
3. **Define** the $N$-th roots of unity $\omega^k$ with $\omega = e^{2\pi i/N}$ and **prove** their summation property $\sum_{k=0}^{N-1}\omega^{k}=0$.
4. **Explain** why quantum amplitudes must be complex — that *relative phase* is physically observable through interference — and connect this to the Fourier machinery developed later in the course.
5. **Compute and visualize** roots of unity in NumPy and verify their algebraic identities numerically.
---
## Intuition
A real number is a point on a line; a complex number is a point in a plane. That extra dimension is
not a bookkeeping convenience — it is *rotation*. Multiplying by $i$ rotates the plane by $90°$;
multiplying by $e^{i\theta}$ rotates it by $\theta$. The single most important picture to carry
forward is this:
> **A unit-modulus complex number $e^{i\theta}$ is a pure rotation by angle $\theta$.**
Why does quantum mechanics need this and not just real probabilities? Consider two paths an
amplitude can take to the same outcome. Classically you add *probabilities*: $p = p_1 + p_2 \ge 0$,
and more paths can only help. Quantum mechanically you add *amplitudes*: $\alpha = \alpha_1 +
\alpha_2$, and then square. If $\alpha_1 = \tfrac12$ and $\alpha_2 = -\tfrac12$, the outcome
probability is $|\alpha_1+\alpha_2|^2 = 0$: the two paths *cancel*. That sign — more generally, that
relative phase $e^{i\theta}$ — is the whole game. Real numbers give you only the signs $\pm 1$;
complex numbers give you a continuous dial of cancellation. We forward-reference this as
**interference**, the engine of every quantum speedup.
Hold two facts together as you read the theory: complex numbers are (1) the algebra of rotations and
(2) the only number system rich enough to let amplitudes interfere continuously while still
collapsing to real probabilities via $|\cdot|^2$.
---
## Theory
### The complex field
**Definition (complex numbers).** The set $\mathbb{C} = \{a + bi : a,b\in\mathbb{R}\}$ with the
symbol $i$ satisfying $i^2 = -1$, equipped with
$$
(a+bi) + (c+di) = (a+c) + (b+d)i,\qquad
(a+bi)(c+di) = (ac-bd) + (ad+bc)i,
$$
forms a **field**: addition and multiplication are commutative, associative, and distributive;
$0 = 0+0i$ and $1 = 1+0i$ are the identities; and every nonzero $z$ has an inverse (derived below).
We call $a = \operatorname{Re}(z)$ the **real part** and $b = \operatorname{Im}(z)$ the **imaginary
part**. (We freely write $i$ where Python writes `1j`; see [Appendix B §1](../../appendix/B-python-numpy-refresher.md).)
**Definition (conjugate and modulus).** The **complex conjugate** of $z = a+bi$ is
$\bar z = z^{*} = a - bi$, and the **modulus** is
$$
|z| = \sqrt{a^2 + b^2} = \sqrt{z\,z^{*}}.
$$
The conjugate flips the sign of the imaginary part (a reflection across the real axis). Two
identities we use constantly:
$$
z\,z^{*} = (a+bi)(a-bi) = a^2 + b^2 = |z|^2 \in \mathbb{R}_{\ge 0},
\qquad
(zw)^{*} = z^{*}w^{*}.
$$
The first gives the multiplicative inverse for $z\neq 0$:
$$
z^{-1} = \frac{z^{*}}{|z|^2} = \frac{a - bi}{a^2+b^2},
$$
which proves $\mathbb{C}$ is a field. Conjugation is an *involution* ($(z^{*})^{*}=z$) and is
additive ($(z+w)^{*}=z^{*}+w^{*}$); together with $(zw)^{*}=z^{*}w^{*}$ it is a *field
automorphism*. This is exactly the operation that turns a ket into a bra: $\langle\psi| =
|\psi\rangle^{\dagger}$ conjugates amplitudes (see [Appendix C §1](../../appendix/C-notation-glossary.md)).
### Cartesian and polar form
Plotting $z=a+bi$ at Cartesian coordinates $(a,b)$ in the **complex plane** (Argand diagram), the
point also has polar coordinates $(r,\theta)$:
$$
a = r\cos\theta,\quad b = r\sin\theta,\qquad
r = |z| = \sqrt{a^2+b^2},\quad \theta = \arg(z) = \operatorname{atan2}(b,a).
$$
We call $r$ the modulus and $\theta$ the **argument**. The argument is only defined modulo $2\pi$;
the **principal value** $\operatorname{Arg}(z)\in(-\pi,\pi]$ is the standard choice (NumPy's
`np.angle` returns it). Use `atan2`, not $\arctan(b/a)$, so the correct quadrant is selected.
### Euler's formula — derivation via power series
We want a meaning for $e^{i\theta}$. Take the real Maclaurin series of $e^x$, $\cos x$, $\sin x$ —
each converges for all real arguments, and the exponential series converges *absolutely* for every
complex argument, so we may extend it to $z\in\mathbb C$ and rearrange freely:
$$
e^{z} = \sum_{n=0}^{\infty}\frac{z^n}{n!},\qquad
\cos x = \sum_{m=0}^{\infty}\frac{(-1)^m x^{2m}}{(2m)!},\qquad
\sin x = \sum_{m=0}^{\infty}\frac{(-1)^m x^{2m+1}}{(2m+1)!}.
$$
Substitute $z = i\theta$ and use the cycle $i^0=1,\ i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1,\dots$:
$$
e^{i\theta} = \sum_{n=0}^{\infty}\frac{(i\theta)^n}{n!}
= \sum_{n=0}^{\infty}\frac{i^n\theta^n}{n!}.
$$
Split into even ($n=2m$, giving $i^{2m}=(-1)^m$) and odd ($n=2m+1$, giving $i^{2m+1}=(-1)^m i$)
terms. Absolute convergence lets us regroup:
$$
e^{i\theta}
= \underbrace{\sum_{m=0}^{\infty}\frac{(-1)^m\theta^{2m}}{(2m)!}}_{=\,\cos\theta}
\;+\; i\underbrace{\sum_{m=0}^{\infty}\frac{(-1)^m\theta^{2m+1}}{(2m+1)!}}_{=\,\sin\theta}.
$$
This is **Euler's formula**:
$$
\boxed{\,e^{i\theta} = \cos\theta + i\sin\theta\,}.
$$
Immediate consequences:
- $|e^{i\theta}| = \sqrt{\cos^2\theta+\sin^2\theta} = 1$ — the unit circle, as the intuition promised.
- $\theta=\pi$ gives **Euler's identity** $e^{i\pi}+1 = 0$.
- $(e^{i\theta})^{*} = \cos\theta - i\sin\theta = e^{-i\theta}$, hence the inverse-on-the-circle:
$e^{i\theta}e^{-i\theta}=1$.
- The **complex exponential** of a general $z=a+bi$ factors a stretch and a rotation:
$$
e^{z} = e^{a+bi} = e^{a}\,e^{ib} = e^{a}\big(\cos b + i\sin b\big).
$$
We can now read **polar form** as an exponential:
$$
z = r(\cos\theta + i\sin\theta) = r\,e^{i\theta},\qquad r=|z|,\ \theta=\arg z.
$$
Multiplication becomes transparent — moduli multiply, arguments add:
$$
(r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = r_1 r_2\, e^{i(\theta_1+\theta_2)}.
$$
This is the precise sense in which *multiplying by $e^{i\theta}$ rotates by $\theta$*.
### De Moivre's theorem
Setting $r=1$ and iterating the multiplication rule (or, formally, by induction on $n$ using
$e^{i(n+1)\theta}=e^{in\theta}e^{i\theta}$) gives **de Moivre's theorem**:
$$
\big(\cos\theta + i\sin\theta\big)^{n} = \cos n\theta + i\sin n\theta,\qquad n\in\mathbb{Z}.
$$
It is just $\big(e^{i\theta}\big)^{n} = e^{in\theta}$ in disguise. It is the workhorse for deriving
multiple-angle identities and, next, the roots of unity.
### The $N$-th roots of unity
We want all solutions of $z^N = 1$ for a positive integer $N$. Write $z = r e^{i\theta}$. Then
$z^N = r^N e^{iN\theta} = 1 = 1\cdot e^{i\cdot 0}$. Matching moduli forces $r=1$; matching arguments
(modulo $2\pi$) forces $N\theta = 2\pi k$ for integer $k$, i.e. $\theta = 2\pi k/N$. Distinct values
arise only for $k=0,1,\dots,N-1$. Defining the **primitive $N$-th root of unity**
$$
\boxed{\,\omega := e^{2\pi i/N}\,}\qquad(\text{matches }[\text{Appendix E §6}](../../appendix/E-math-identities.md)),
$$
the $N$ roots are exactly the powers of $\omega$:
$$
z_k = \omega^{k} = e^{2\pi i k/N},\qquad k = 0,1,\dots,N-1.
$$
Geometrically they are the $N$ vertices of a regular $N$-gon inscribed in the unit circle, starting
at $1$ and stepping counter-clockwise by $2\pi/N$. (We use the sign convention $\omega =
e^{+2\pi i/N}$ throughout the program; see [Appendix C §4](../../appendix/C-notation-glossary.md). The
DFT in lesson 3 uses $e^{-2\pi i/N}$ as its forward kernel — a deliberate, consistent choice.)
**Theorem (vanishing sum).** For $N>1$,
$$
\sum_{k=0}^{N-1}\omega^{k} = 0.
$$
*Proof.* The sum is a finite geometric series with ratio $\omega$. Since $N>1$, $\omega = e^{2\pi
i/N}\neq 1$ (its argument $2\pi/N\in(0,2\pi)$ is nonzero mod $2\pi$), so the closed form applies:
$$
\sum_{k=0}^{N-1}\omega^{k} = \frac{1-\omega^{N}}{1-\omega}.
$$
But $\omega^{N} = e^{2\pi i} = 1$, so the numerator is $1-1 = 0$ while the denominator $1-\omega\neq
0$. Hence the sum is $0$. $\blacksquare$
This is the algebraic heart of all Fourier analysis: distinct frequency components *cancel* when
summed over a full period. We generalize it now.
**Corollary (orthogonality of roots).** For integers $j,k$,
$$
\frac{1}{N}\sum_{x=0}^{N-1} \omega^{x(j-k)} = \delta_{jk}
\qquad\text{(indices taken mod }N),
$$
matching the DFT/QFT orthogonality in [Appendix E §6](../../appendix/E-math-identities.md).
*Proof.* Let $m = j-k$. If $m\equiv 0\pmod N$ then every term is $\omega^{0}=1$ and the average is
$\tfrac1N\cdot N = 1$. Otherwise $\omega^{m}\neq 1$ and the same geometric-series argument gives
$$
\sum_{x=0}^{N-1}(\omega^{m})^{x} = \frac{1-(\omega^{m})^{N}}{1-\omega^{m}}
= \frac{1-(\omega^{N})^{m}}{1-\omega^{m}} = \frac{1-1}{1-\omega^{m}} = 0. \qquad\blacksquare
$$
**Other useful properties.**
- **Closure / group structure.** The roots $\{1,\omega,\dots,\omega^{N-1}\}$ are closed under
multiplication: $\omega^{j}\omega^{k} = \omega^{(j+k)\bmod N}$. They form the cyclic group
$\mathbb Z/N\mathbb Z$ under addition of exponents — the same group that indexes computational
basis states in the QFT.
- **Conjugate = inverse.** $\overline{\omega^{k}} = \omega^{-k} = \omega^{N-k}$, since
$|\omega^k|=1$.
- **Product of all roots.** $\prod_{k=0}^{N-1}\omega^{k} = \omega^{\,0+1+\cdots+(N-1)} =
\omega^{N(N-1)/2} = (-1)^{N-1}$.
### Why quantum amplitudes are complex
A pure state is $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$ with $\alpha,\beta\in\mathbb C$ and
$|\alpha|^2 + |\beta|^2 = 1$ ([Appendix C §1](../../appendix/C-notation-glossary.md)). Measurement
gives outcome $0$ with probability $|\alpha|^2$ — so why not just store the probabilities
$(|\alpha|^2,|\beta|^2)$ and stay real and nonnegative?
Because *evolution mixes the components*, and the mixing is sensitive to phase. Apply a Hadamard to
$|+\rangle = \tfrac1{\sqrt2}(|0\rangle+|1\rangle)$ versus $|-\rangle =
\tfrac1{\sqrt2}(|0\rangle-|1\rangle)$:
$$
H|+\rangle = |0\rangle,\qquad H|-\rangle = |1\rangle.
$$
Both $|+\rangle$ and $|-\rangle$ have the *same* measurement statistics in the computational basis
($50/50$), so the probability vector $(\tfrac12,\tfrac12)$ cannot tell them apart. Yet $H$ sends one
to a certain $0$ and the other to a certain $1$. The only thing distinguishing them is the
**relative phase** $\beta/\alpha = +1$ versus $-1$ — a phase $e^{i\pi}$. Relative phase is therefore
*physically observable*: it changes future measurement outcomes after interference.
Two more points pin this down:
- **Global phase is unobservable.** $|\psi\rangle$ and $e^{i\gamma}|\psi\rangle$ give identical
statistics for *every* measurement, because $|e^{i\gamma}\alpha|^2 = |\alpha|^2$. Only phase
*differences between amplitudes* matter — exactly the relative phase above.
- **Why not just signs $\pm1$?** Real amplitudes ($\beta/\alpha\in\{\pm1\}$ up to scale) allow only
two-way cancellation. Continuous phases $e^{i\theta}$ let amplitudes interfere by *any* angle, and
the canonical phases are precisely the roots of unity $\omega^k$. When a quantum register holds a
uniform superposition and we apply a phase $\omega^{xy}$ to basis state $|y\rangle$, the
interference pattern is governed by sums like $\sum_y \omega^{xy}$ — which, by the vanishing-sum
theorem above, are zero except at special $x$. That selective cancellation *is* the Quantum
Fourier Transform, developed in [Term 2.4 · The Quantum Fourier Transform](../../term-2-qc-core/c04-qft-qpe/01-quantum-fourier-transform.md).
So: amplitudes are complex because nature lets probability amplitudes *rotate and interfere*, and
$\mathbb C$ is the minimal arena — a field where $|\cdot|^2$ recovers probabilities — in which
continuous rotation lives.
---
## Worked Examples
### Example 1 — Polar form, powers, and a root
Let $z = 1 + i\sqrt 3$.
**Modulus and argument.** $|z| = \sqrt{1^2 + (\sqrt3)^2} = \sqrt{4} = 2$. The point $(1,\sqrt3)$ is
in the first quadrant, so $\arg z = \operatorname{atan2}(\sqrt3, 1) = \pi/3$. Thus
$$
z = 2\,e^{i\pi/3}.
$$
**A power, via de Moivre.** Compute $z^{6}$:
$$
z^{6} = 2^{6}\,e^{i\cdot 6\pi/3} = 64\,e^{i2\pi} = 64\cdot 1 = 64.
$$
The phase wraps exactly once around the circle and lands back on the positive real axis — a clean
illustration of "arguments add (mod $2\pi$), moduli multiply."
**A square root.** $\sqrt z = \sqrt2\,e^{i\pi/6} = \sqrt2(\cos\tfrac\pi6 + i\sin\tfrac\pi6) =
\sqrt2\big(\tfrac{\sqrt3}{2} + \tfrac{i}{2}\big) = \tfrac{\sqrt6}{2} + \tfrac{\sqrt2}{2}\,i$. (The
other root is its negative.)
### Example 2 — The cube roots of unity and their sum
Take $N=3$, $\omega = e^{2\pi i/3}$. The roots are:
$$
\omega^{0} = 1,\qquad
\omega^{1} = e^{2\pi i/3} = \cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3} = -\tfrac12 + \tfrac{\sqrt3}{2}i,
$$
$$
\omega^{2} = e^{4\pi i/3} = -\tfrac12 - \tfrac{\sqrt3}{2}i.
$$
They are the vertices of an equilateral triangle on the unit circle. Their sum:
$$
1 + \Big(-\tfrac12 + \tfrac{\sqrt3}{2}i\Big) + \Big(-\tfrac12 - \tfrac{\sqrt3}{2}i\Big)
= (1 - \tfrac12 - \tfrac12) + \big(\tfrac{\sqrt3}{2} - \tfrac{\sqrt3}{2}\big)i = 0,
$$
confirming $\sum_{k=0}^{2}\omega^k = 0$. Note also $\omega^2 = \overline{\omega^1} = \omega^{-1}$, as
the conjugate-inverse property predicts.
### Example 3 — Interference: distinguishing $|+\rangle$ from $|-\rangle$
Let $\alpha = \tfrac1{\sqrt2}$ and consider $|\pm\rangle = \alpha|0\rangle \pm \alpha|1\rangle$. The
Hadamard $H = \tfrac1{\sqrt2}\begin{psmallmatrix}1&1\\1&-1\end{psmallmatrix}$ acts as
$$
H|+\rangle = \tfrac12\big[(1+1)|0\rangle + (1-1)|1\rangle\big] = |0\rangle,
$$
$$
H|-\rangle = \tfrac12\big[(1-1)|0\rangle + (1+1)|1\rangle\big] = |1\rangle.
$$
In $H|+\rangle$ the $|1\rangle$ amplitudes $\tfrac12$ and $-\tfrac12$ cancel by *destructive
interference* (their relative phase is $e^{i\pi}=-1$); in $H|-\rangle$ the $|0\rangle$ amplitudes
cancel instead. A purely real, nonnegative probability description $(\tfrac12,\tfrac12)$ for both
inputs could never predict these opposite, deterministic outputs. The complex (here, just signed)
relative phase is doing physical work — this is the smallest possible instance of the interference
that the Fourier methods in this course exploit at scale.
The same two lines describe a Mach–Zehnder interferometer: a beam splitter $H$, a phase
$e^{i\varphi}$ on one arm, and a second $H$. At $\varphi = 0$ the photon is in $|+\rangle$ between the
splitters and always exits to detector $D_0$. At $\varphi = \pi$ it is in $|-\rangle$ and always exits
to $D_1$. In between, $D_0$ fires with probability $\cos^2(\varphi/2)$, the result of E6 below:
```qm-demo mach-zehnder
```
---
## Hands-on (Python)
```python
import numpy as np
import matplotlib.pyplot as plt
# ----- Cartesian <-> polar -----
z = 1 + 1j * np.sqrt(3) # 1 + i*sqrt(3)
print("z =", z)
print("|z| =", abs(z)) # 2.0 = modulus
print("arg(z) [rad] =", np.angle(z)) # ~1.047 = pi/3
print("conj(z) =", z.conjugate()) # 1 - i*sqrt(3)
print("z * conj(z) =", (z * z.conjugate()).real) # |z|^2 = 4.0
# Reconstruct z from polar form r * e^{i*theta}
r, theta = abs(z), np.angle(z)
z_polar = r * np.exp(1j * theta)
print("polar matches Cartesian:", np.isclose(z, z_polar)) # True
def roots_of_unity(N):
"""Return the N-th roots of unity as a complex array [omega^0, ..., omega^{N-1}]."""
k = np.arange(N)
omega = np.exp(2j * np.pi / N) # primitive root ω = e^{2πi/N}
return omega ** k # equivalently np.exp(2j*np.pi*k/N)
# ----- Verify the vanishing-sum theorem Σ ω^k = 0 for several N -----
for N in (2, 3, 4, 8, 16):
s = roots_of_unity(N).sum()
print(f"N={N:2d}: Σ ω^k = {s:.2e} (|sum| = {abs(s):.2e})")
assert np.isclose(s, 0.0, atol=1e-12), "vanishing-sum theorem failed!"
# ----- Verify orthogonality (1/N) Σ_x ω^{x(j-k)} = δ_{jk} -----
N = 8
omega = np.exp(2j * np.pi / N)
x = np.arange(N)
# Build the N x N matrix M[j, k] = (1/N) Σ_x ω^{x(j-k)}; it should be the identity.
M = np.array([[np.mean(omega ** (x * (j - k))) for k in range(N)] for j in range(N)])
print("orthogonality gives identity:", np.allclose(M, np.eye(N), atol=1e-12)) # True
def plot_roots_of_unity(N):
"""Plot the N-th roots of unity on the unit circle."""
roots = roots_of_unity(N)
fig, ax = plt.subplots(figsize=(5, 5))
circle = plt.Circle((0, 0), 1.0, fill=False, linestyle="--", color="gray")
ax.add_patch(circle)
ax.scatter(roots.real, roots.imag, zorder=3)
for k, w in enumerate(roots):
ax.annotate(rf"$\omega^{{{k}}}$", (w.real, w.imag),
textcoords="offset points", xytext=(6, 6))
ax.axhline(0, color="black", lw=0.5); ax.axvline(0, color="black", lw=0.5)
ax.set_aspect("equal"); ax.set_xlim(-1.3, 1.3); ax.set_ylim(-1.3, 1.3)
ax.set_title(rf"$N={N}$ roots of unity, $\omega=e^{{2\pi i/{N}}}$")
ax.set_xlabel("Re"); ax.set_ylabel("Im")
plt.tight_layout(); plt.show()
plot_roots_of_unity(8)
# ----- Interference demo: H distinguishes |+> from |-> -----
H = np.array([[1, 1], [1, -1]], dtype=complex) / np.sqrt(2)
ket_plus = np.array([1, 1], dtype=complex) / np.sqrt(2)
ket_minus = np.array([1, -1], dtype=complex) / np.sqrt(2)
print("H|+> =", np.round(H @ ket_plus, 12).real) # [1, 0] -> |0>
print("H|-> =", np.round(H @ ket_minus, 12).real) # [0, 1] -> |1>
```
Running this prints `|z|=2.0`, `arg(z)≈1.047` ($=\pi/3$), confirms $\sum\omega^k=0$ to machine
precision for every $N$, verifies the orthogonality matrix is the identity, draws the regular
$8$-gon of roots, and shows that $H$ maps $|+\rangle\mapsto|0\rangle$ and
$|-\rangle\mapsto|1\rangle$ — interference made visible.
---
## Exercises
Attempt each before expanding the solution.
**1. (Easy) Cartesian ↔ polar.** Write $z = -\sqrt3 + i$ in polar form $re^{i\theta}$ with
$\theta\in(-\pi,\pi]$, then compute $z^{4}$.
Solution
$|z| = \sqrt{(\sqrt3)^2 + 1^2} = 2$. The point $(-\sqrt3, 1)$ is in the second quadrant:
$\theta = \operatorname{atan2}(1, -\sqrt3) = \pi - \pi/6 = 5\pi/6$. So $z = 2\,e^{i5\pi/6}$.
Then $z^4 = 2^4 e^{i\cdot 20\pi/6} = 16\,e^{i10\pi/3}$. Reduce the argument mod $2\pi$:
$10\pi/3 - 2\pi = 4\pi/3$, so $z^4 = 16\,e^{i4\pi/3} = 16(-\tfrac12 - \tfrac{\sqrt3}{2}i)
= -8 - 8\sqrt3\,i$.
Solution
$zw = (ac-bd) + (ad+bc)i$, so $(zw)^{*} = (ac-bd) - (ad+bc)i$. Meanwhile $z^{*}w^{*} =
(a-bi)(c-di) = (ac - bd) - (ad+bc)i$ (the cross terms $-adi - bci$ combine to $-(ad+bc)i$, and
$(-bi)(-di) = bd i^2 = -bd$). They match. Then $|zw|^2 = (zw)(zw)^{*} = z w z^{*} w^{*} =
(zz^{*})(ww^{*}) = |z|^2|w|^2$; taking nonnegative square roots gives $|zw| = |z||w|$. (In polar
form this is just "moduli multiply.")
Solution
$(\cos\theta + i\sin\theta)^3 = \cos3\theta + i\sin3\theta$. Expand the left side with the binomial
theorem: $\cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3
= \cos^3\theta - 3\cos\theta\sin^2\theta + i(3\cos^2\theta\sin\theta - \sin^3\theta)$. Match real
parts: $\cos3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta$. Substitute $\sin^2\theta = 1 -
\cos^2\theta$: $\cos3\theta = \cos^3\theta - 3\cos\theta(1-\cos^2\theta) = 4\cos^3\theta -
3\cos\theta$. $\blacksquare$
Solution
$\omega^{2} = e^{4\pi i/6} = e^{2\pi i/3}$ is a primitive *cube* root of unity. As $k$ runs over
$0,\dots,5$, $\omega^{2k}$ runs over the cube roots $\{1, e^{2\pi i/3}, e^{4\pi i/3}\}$ *twice*
(period 3). So $\sum_{k=0}^{5}\omega^{2k} = 2\sum_{j=0}^{2}(e^{2\pi i/3})^{j} = 2\cdot 0 = 0$ by the
vanishing-sum theorem for $N=3$. Group-theoretically, $\{\omega^{2k}\}$ is the subgroup generated by
$\omega^2$, which is the cyclic group of order $3$; the sum over any nontrivial finite cyclic
subgroup (here covered twice) vanishes.
Solution
Sum the geometric series with ratio $e^{i\phi}\neq 1$:
$S = \sum_{k=0}^{N-1} e^{ik\phi} = \frac{1 - e^{iN\phi}}{1 - e^{i\phi}}$. Factor out the
half-angle from numerator and denominator ("$1-e^{i\alpha} = e^{i\alpha/2}(e^{-i\alpha/2} -
e^{i\alpha/2}) = -2i\,e^{i\alpha/2}\sin(\alpha/2)$"):
$$
S = \frac{-2i\,e^{iN\phi/2}\sin(N\phi/2)}{-2i\,e^{i\phi/2}\sin(\phi/2)}
= e^{i(N-1)\phi/2}\,\frac{\sin(N\phi/2)}{\sin(\phi/2)}.
$$
Taking the modulus (the leading exponential has modulus 1) gives
$|S| = \big|\sin(N\phi/2)/\sin(\phi/2)\big|$. For $\phi = 2\pi/N$: $N\phi/2 = \pi$, so
$\sin(N\phi/2) = \sin\pi = 0$ while $\sin(\phi/2) = \sin(\pi/N)\neq 0$ for $N>1$. Hence $|S|=0$,
i.e. $\sum_k \omega^k = 0$. $\blacksquare$
Solution
(a) $|\langle 0|\psi_\varphi\rangle|^2 = |\tfrac1{\sqrt2}|^2 = \tfrac12$ and likewise for outcome
$1$, independent of $\varphi$ — the global phase on $|1\rangle$ is washed out by $|\cdot|^2$.
(b) $H|\psi_\varphi\rangle = \tfrac1{\sqrt2}\big[\tfrac1{\sqrt2}(|0\rangle+|1\rangle) + e^{2\pi
i\varphi}\tfrac1{\sqrt2}(|0\rangle-|1\rangle)\big] = \tfrac12\big[(1+e^{2\pi i\varphi})|0\rangle +
(1-e^{2\pi i\varphi})|1\rangle\big]$. The amplitude of $|0\rangle$ is $\tfrac12(1+e^{2\pi
i\varphi})$. Using $1+e^{i\alpha} = e^{i\alpha/2}(e^{-i\alpha/2}+e^{i\alpha/2}) = 2e^{i\alpha/2}
\cos(\alpha/2)$ with $\alpha = 2\pi\varphi$, the amplitude is $e^{i\pi\varphi}\cos(\pi\varphi)$, so
$p(0) = |{\cdot}|^2 = \cos^2(\pi\varphi)$. The previously hidden phase $\varphi$ now controls a
measurable probability — interference has converted *relative phase* into *statistics*. This is
exactly the mechanism scaled up in [Term 2.4](../../term-2-qc-core/c04-qft-qpe/01-quantum-fourier-transform.md).
1. State Euler's formula and give the one-line reason $|e^{i\theta}|=1$.
$e^{i\theta} = \cos\theta + i\sin\theta$. Its modulus is $\sqrt{\cos^2\theta+\sin^2\theta}=1$, so it
lies on the unit circle — a pure rotation by $\theta$.
2. Why does $\sum_{k=0}^{N-1}\omega^k = 0$ for $N>1$, in one sentence?
It is a geometric series with ratio $\omega\neq1$; its closed form $(1-\omega^N)/(1-\omega)$ has
numerator $1-\omega^N = 1-1 = 0$ because $\omega^N = e^{2\pi i} = 1$.
3. What is the difference between global phase and relative phase, and which is physical?
A global phase $e^{i\gamma}$ multiplies the *whole* state and is unobservable ($|e^{i\gamma}\alpha|^2
= |\alpha|^2$). A relative phase is a phase *difference between amplitudes* (e.g. $\beta/\alpha$); it
is physically observable because it changes interference and hence future measurement statistics.
4. Concretely, why can't real nonnegative probabilities replace complex amplitudes?
Because evolution mixes amplitudes, and the result depends on phases. $|+\rangle$ and $|-\rangle$
share the probability vector $(\tfrac12,\tfrac12)$ but $H$ maps them to $|0\rangle$ and $|1\rangle$
respectively — opposite deterministic outcomes. Only the complex (relative) phase distinguishes the
inputs.
5. How are the $N$-th roots of unity related to the group $\mathbb Z/N\mathbb Z$?
$\omega^j\omega^k = \omega^{(j+k)\bmod N}$, so the roots are closed under multiplication and form a
cyclic group isomorphic to $(\mathbb Z/N\mathbb Z, +)$ via $k\mapsto\omega^k$. This is the group that
indexes basis states in the (Quantum) Fourier Transform.
6. What single identity from this lesson is the algebraic core of the DFT/QFT?
The orthogonality relation $\tfrac1N\sum_{x=0}^{N-1}\omega^{x(j-k)} = \delta_{jk}$ — distinct
frequency components cancel over a full period, leaving only the matching term. We use it to prove
the DFT matrix is unitary in [lesson 3](03-discrete-fourier-transform.md).