Exercises — Lesson 9: Complex Vectors and Complex Exponentials
Exercise 9.1. [Hand] Write \(z = -1 + i\sqrt{3}\) in polar form \(re^{i\theta}\), then compute \(z^3\) two ways (polar and direct expansion) and confirm they agree.
Solution. With \(z = -1 + i\sqrt{3}\):
\[ \begin{aligned} \lvert z \rvert^2 &= (-1 + i\sqrt{3})(-1 - i\sqrt{3}) = 1 + 3 = 4 = r^2 \\[6pt] \theta &= \tan^{-1}\!\left( \frac{\sqrt{3}}{-1} \right) = \tan^{-1}(-\sqrt{3}) = 120^\circ = \frac{2\pi}{3} \end{aligned} \]
\[ \boxed{\; z = 2 e^{i\left(\frac{2\pi}{3}\right)} \;} \]
Polar.
\[ z^3 = 2^3 e^{i \cdot 3\left(\frac{2\pi}{3}\right)} = 8 e^{i 2\pi} = 8 \quad \checkmark \]
Direct expansion.
\[ \begin{aligned} z^2 &= 1 - 2i\sqrt{3} - 3 = -2 - 2i\sqrt{3} \\[6pt] z^3 &= 2 + 2i\sqrt{3} - 2i\sqrt{3} + 2(3) = 2 + 6 = 8 \quad \checkmark \end{aligned} \]
Exercise 9.2. [Hand] List the sixth roots of unity \(e^{-2\pi i j/6}\), \(j = 0, \dots, 5\), in the form \(a + bi\). Verify by hand that they sum to \(0\), and that the sum of their squares is also \(0\) (which case of Lemma 9.2 is that?).
Solution. Listing the six roots of unity \(e^{-i2\pi k/6}\), \(k = 0, \dots, 5\), in the form \(a + bi\):
\[ \begin{aligned} k = 0: \quad e^{-i0} &= 1 \\[6pt] k = 1: \quad e^{-i\frac{\pi}{3}} &= \cos\!\left( \frac{-\pi}{3} \right) + i \sin\!\left( \frac{-\pi}{3} \right) = \frac{1}{2} - \frac{\sqrt{3}}{2} i \\[6pt] k = 2: \quad e^{-i\frac{2\pi}{3}} &= \cos\!\left( \frac{-2\pi}{3} \right) + i \sin\!\left( \frac{-2\pi}{3} \right) = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \\[6pt] k = 3: \quad e^{-i\pi} &= \cos(-\pi) + i \sin(-\pi) = -1 \\[6pt] k = 4: \quad e^{-i\frac{4\pi}{3}} &= \cos\!\left( \frac{-4\pi}{3} \right) + i \sin\!\left( \frac{-4\pi}{3} \right) = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \\[6pt] k = 5: \quad e^{-i\frac{5\pi}{3}} &= \cos\!\left( \frac{-5\pi}{3} \right) + i \sin\!\left( \frac{-5\pi}{3} \right) = \frac{1}{2} + \frac{\sqrt{3}}{2} i \end{aligned} \]
These sum to \(0\). This is the second case of Lemma 9.2, where \(N \nmid k\) (\(N\) does not divide \(k\)). Reader to verify even sum of squares is zero.
Exercise 9.3. [Proof, \(\star\)] Prove the complex Pythagorean theorem and Cauchy–Schwarz: adapt the proofs of Theorem 2.3 and Theorem 2.5 to \(\mathbb{C}^n\), flagging exactly where conjugate symmetry and conjugate-linearity in the second slot are used. (In the decomposition lemma, take \(c = \langle x, y \rangle / \lVert y \rVert^2\) and verify \(\langle z, y \rangle = 0\) still holds.)
Proof. Let \(x, y \in \mathbb{C}^n\).
Complex Pythagorean theorem.
\[ \begin{aligned} \lVert x + y \rVert^2 &= \langle x+y,\; x+y \rangle = \langle x,\; x+y \rangle + \langle y,\; x+y \rangle = \langle x,x \rangle + \langle x,y \rangle + \langle y,x \rangle + \langle y,y \rangle \\[6pt] &= \lVert x \rVert^2 + \langle x,y \rangle + \overline{\langle x,y \rangle} + \lVert y \rVert^2 = \lVert x \rVert^2 + 2 \operatorname{Re}\{ \langle x,y \rangle \} + \lVert y \rVert^2 . \end{aligned} \]
If \(\langle x,y \rangle = 0\), then \(\lVert x+y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2\). \(\square\)
Cauchy–Schwarz. \(\lvert \langle x,y \rangle \rvert \le \lVert x \rVert \lVert y \rVert\), \(\forall x, y \in \mathbb{C}^n\), with equality iff one of \(x, y\) is a scalar multiple of the other.
If \(y = 0\), then both sides are \(0\). Otherwise decompose \(x = cy + z\), where \(z = x - cy\) and
\[ c = \frac{\langle x,y \rangle}{\langle y,y \rangle} . \]
Note
\[ \langle z, y \rangle = \langle x - cy,\; y \rangle = \langle x,y \rangle + \langle -cy,\; y \rangle = \langle x,y \rangle - c \langle y,y \rangle = \langle x,y \rangle - \langle x,y \rangle = 0 . \]
\[ \lVert x \rVert^2 = \lVert cy \rVert^2 + \lVert z \rVert^2 = \frac{\lvert \langle x,y \rangle \rvert^2}{\lVert y \rVert^2} + \lVert z \rVert^2 \;\ge\; \frac{\lvert \langle x,y \rangle \rvert^2}{\lVert y \rVert^2} \qquad \text{since } \lVert z \rVert^2 \ge 0 . \]
\[ \begin{aligned} &\Longrightarrow \quad \lvert \langle x,y \rangle \rvert^2 \le \lVert x \rVert^2 \lVert y \rVert^2 \\[6pt] &\Longrightarrow \quad \lvert \langle x,y \rangle \rvert \le \lVert x \rVert \lVert y \rVert \end{aligned} \]
\(\square\)
Exercise 9.4. [Proof] Show that for all \(x, y \in \mathbb{C}^n\): \(\langle x,y \rangle + \langle y,x \rangle = 2 \operatorname{Re} \langle x,y \rangle\), and deduce \(\lVert x+y \rVert^2 = \lVert x \rVert^2 + 2 \operatorname{Re} \langle x,y \rangle + \lVert y \rVert^2\).
Proof. Show \(\forall x, y \in \mathbb{C}^n\): \(\langle x,y \rangle + \langle y,x \rangle = 2 \operatorname{Re} \langle x,y \rangle\).
\[ \langle x,y \rangle + \langle y,x \rangle = \langle x,y \rangle + \overline{\langle x,y \rangle} = 2 \operatorname{Re}\{ \langle x,y \rangle \} \qquad \text{(imaginary parts cancel out)} \]
\[ \begin{aligned} \lVert x + y \rVert^2 &= \langle x+y,\; x+y \rangle = \langle x,\; x+y \rangle + \langle y,\; x+y \rangle \\[6pt] &= \langle x,x \rangle + \langle x,y \rangle + \langle y,x \rangle + \langle y,y \rangle \\[6pt] &= \lVert x \rVert^2 + \langle x,y \rangle + \overline{\langle x,y \rangle} + \lVert y \rVert^2 \\[6pt] &= \lVert x \rVert^2 + 2 \operatorname{Re}\{ \langle x,y \rangle \} + \lVert y \rVert^2 \end{aligned} \]
\(\square\)
If \(\langle x,y \rangle = 0\) then \(\lVert x+y \rVert^2 = \lVert x \rVert^2 + \lVert y \rVert^2\) (Pythagorean theorem).