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Appendix A Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Multiplication of Complex Numbers

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(a) \((2 + 3i)(4 - i)\)

Strategy: Expand normally and substitute \(i^2 = -1\).

\[ (2 + 3i)(4 - i) = 2 \cdot 4 + 2 \cdot (-i) + 3i \cdot 4 + 3i \cdot (-i) \]
\[ = 8 - 2i + 12i - 3i^2 = 8 + 10i - 3(-1) = 8 + 10i + 3 \]
\[ \boxed{11 + 10i} \]

Verification: Check using the product of absolute values. \(|2+3i|^2 = 4+9 = 13\), \(|4-i|^2 = 16+1 = 17\). The squared absolute value of the product is \(13 \times 17 = 221\). For the result, \(|11+10i|^2 = 121 + 100 = 221\). ✓

(b) \((1 + i)^3\)

First compute \((1+i)^2\).

\[ (1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i \]

Then multiply by \((1+i)\).

\[ (1+i)^3 = 2i \cdot (1+i) = 2i + 2i^2 = 2i - 2 \]
\[ \boxed{-2 + 2i} \]

Verification: \(|1+i|^2 = 2\) so \(|1+i|^6 = 8\). \(|-2+2i|^2 = 4+4 = 8\). ✓

(c) \((-2 + i)(3 + 2i)\)

\[ (-2+i)(3+2i) = -6 - 4i + 3i + 2i^2 = -6 - i + 2(-1) = -6 - i - 2 \]
\[ \boxed{-8 - i} \]

Verification: \(|-2+i|^2 = 4+1=5\), \(|3+2i|^2 = 9+4=13\). Product \(= 65\). \(|-8-i|^2 = 64+1=65\). ✓

B-2. Division of Complex Numbers

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(a) \(\dfrac{2 + i}{1 + 3i}\)

Strategy: Multiply the numerator and denominator by the complex conjugate of the denominator, \(1 - 3i\), to make the denominator real.

\[ \frac{2+i}{1+3i} \cdot \frac{1-3i}{1-3i} = \frac{(2+i)(1-3i)}{(1+3i)(1-3i)} \]

Denominator: \((1+3i)(1-3i) = 1 + 9 = 10\)

Numerator: \((2+i)(1-3i) = 2 - 6i + i - 3i^2 = 2 - 5i + 3 = 5 - 5i\)

\[ \boxed{\frac{1}{2} - \frac{1}{2}i} \]

Verification: \(\left(\frac{1}{2} - \frac{1}{2}i\right)(1+3i) = \frac{1}{2} + \frac{3}{2}i - \frac{1}{2}i - \frac{3}{2}i^2 = \frac{1}{2} + i + \frac{3}{2} = 2 + i\). ✓

(b) \(\dfrac{5}{2 - i}\)

\[ \frac{5}{2-i} \cdot \frac{2+i}{2+i} = \frac{5(2+i)}{4+1} = \frac{10+5i}{5} \]
\[ \boxed{2 + i} \]

Verification: \((2+i)(2-i) = 4+1 = 5\). ✓

(c) \(\dfrac{1 + 2i}{3 - 4i}\)

\[ \frac{1+2i}{3-4i} \cdot \frac{3+4i}{3+4i} = \frac{(1+2i)(3+4i)}{9+16} \]

Numerator: \((1+2i)(3+4i) = 3 + 4i + 6i + 8i^2 = 3 + 10i - 8 = -5 + 10i\)

\[ \frac{-5+10i}{25} = \boxed{-\frac{1}{5} + \frac{2}{5}i} \]

Verification: \(\left(-\frac{1}{5}+\frac{2}{5}i\right)(3-4i) = -\frac{3}{5}+\frac{4}{5}i+\frac{6}{5}i-\frac{8}{5}i^2 = -\frac{3}{5}+\frac{10}{5}i+\frac{8}{5} = \frac{5}{5}+2i = 1+2i\). ✓

B-3. Absolute Value and Complex Conjugate

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(a) \(z = 5 - 12i\)

\[ z^* = 5 + 12i, \qquad |z| = \sqrt{25 + 144} = \sqrt{169} = \boxed{13} \]

Check: \(zz^* = (5-12i)(5+12i) = 25 + 144 = 169 = 13^2 = |z|^2\). ✓

(b) \(z = -3i\)

\[ z^* = 3i, \qquad |z| = \sqrt{0 + 9} = \boxed{3} \]

Check: \(zz^* = (-3i)(3i) = -9i^2 = 9 = 3^2\). ✓

(c) \(z = -2 + 2i\)

\[ z^* = -2 - 2i, \qquad |z| = \sqrt{4 + 4} = \sqrt{8} = \boxed{2\sqrt{2}} \]

Check: \(zz^* = (-2+2i)(-2-2i) = 4 + 4 = 8 = (2\sqrt{2})^2\). ✓

(d) \(z = 7\)

\[ z^* = 7, \qquad |z| = \sqrt{49} = \boxed{7} \]

The complex conjugate of a real number is itself. \(zz^* = 49 = |z|^2\). ✓

B-4. Conversion to Polar Form

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(a) \(z = 1 + \sqrt{3}\,i\)

\[ r = \sqrt{1 + 3} = 2 \]

\(\tan\theta = \sqrt{3}/1 = \sqrt{3}\), and since it is in the first quadrant, \(\theta = \pi/3\).

\[ \boxed{z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)} \]

Check: \(2(\cos 60° + i\sin 60°) = 2\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 1 + \sqrt{3}\,i\). ✓

(b) \(z = -2\)

\[ r = 2 \]

Since it points in the negative real axis direction, \(\theta = \pi\).

\[ \boxed{z = 2(\cos\pi + i\sin\pi)} \]

Check: \(2(-1 + 0i) = -2\). ✓

(c) \(z = -1 - i\)

\[ r = \sqrt{1 + 1} = \sqrt{2} \]

With \(a = -1, b = -1\), this is in the third quadrant. \(\tan\theta = (-1)/(-1) = 1\), but since it is in the third quadrant, \(\theta = -\frac{3\pi}{4}\) (in the range \(-\pi < \theta \leq \pi\)).

\[ \boxed{z = \sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i\sin\left(-\frac{3\pi}{4}\right)\right)} \]

Check: \(\cos(-3\pi/4) = -\frac{\sqrt{2}}{2}\), \(\sin(-3\pi/4) = -\frac{\sqrt{2}}{2}\). \(\sqrt{2}\left(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = -1 - i\). ✓

(d) \(z = 3i\)

\[ r = 3 \]

Since it points in the positive imaginary axis direction, \(\theta = \pi/2\).

\[ \boxed{z = 3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)} \]

Check: \(3(0 + i) = 3i\). ✓

B-5. Powers of \(i\)

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Powers of \(i\) cycle with period 4: \(i^0 = 1,\; i^1 = i,\; i^2 = -1,\; i^3 = -i\).

(a) \(i^5\)

Since \(5 = 4 \times 1 + 1\), we have \(i^5 = i^1 = \boxed{i}\)

(b) \(i^{13}\)

Since \(13 = 4 \times 3 + 1\), we have \(i^{13} = i^1 = \boxed{i}\)

(c) \(i^{-1}\)

\(i^{-1} = \dfrac{1}{i} = \dfrac{i}{i^2} = \dfrac{i}{-1} = -i\). Alternatively, \(i^{-1} = i^3 = \boxed{-i}\)

Check: \((-i) \cdot i = -i^2 = 1\). ✓

(d) \(i^{100}\)

Since \(100 = 4 \times 25 + 0\), we have \(i^{100} = i^0 = \boxed{1}\)

B-6. Rules for Complex Conjugation

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\(z_1 = 2 + i\), \(z_2 = 1 - 3i\).

(a) Verification of \((z_1 z_2)^* = z_1^* z_2^*\)

Left-hand side:

\[ z_1 z_2 = (2+i)(1-3i) = 2 - 6i + i - 3i^2 = 2 - 5i + 3 = 5 - 5i \]
\[ (z_1 z_2)^* = 5 + 5i \]

Right-hand side:

\[ z_1^* = 2 - i, \quad z_2^* = 1 + 3i \]
\[ z_1^* z_2^* = (2-i)(1+3i) = 2 + 6i - i - 3i^2 = 2 + 5i + 3 = 5 + 5i \]

LHS \(= 5 + 5i =\) RHS. ✓ \(\quad \blacksquare\)

(b) Verification of \((z_1 + z_2)^* = z_1^* + z_2^*\)

Left-hand side:

\[ z_1 + z_2 = (2+i) + (1-3i) = 3 - 2i \]
\[ (z_1 + z_2)^* = 3 + 2i \]

Right-hand side:

\[ z_1^* + z_2^* = (2-i) + (1+3i) = 3 + 2i \]

LHS \(= 3 + 2i =\) RHS. ✓ \(\quad \blacksquare\)

B-7. Calculating Terms of a Maclaurin Expansion

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Writing out the Maclaurin expansion of \(e^x\) up to the 5th-order term:

\[ e^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} \]

Substituting \(x = 2\):

\[ e^2 \approx 1 + 2 + \frac{4}{2} + \frac{8}{6} + \frac{16}{24} + \frac{32}{120} \]
\[ = 1 + 2 + 2 + 1.3333\ldots + 0.6667\ldots + 0.2667\ldots \]
\[ \approx 7.27 \]
\[ \boxed{e^2 \approx 7.27} \]

Verification: The true value is \(e^2 \approx 7.389\). Adding the 6th-order term \(\frac{2^6}{6!} = \frac{64}{720} \approx 0.089\) gives \(7.36\), which is even closer. This is a reasonable approximation for truncation at 5th order. ✓

B-8. Calculating \(e^{i\theta}\)

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We use Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\).

(a) \(e^{i\pi/4}\)

\[ e^{i\pi/4} = \cos\frac{\pi}{4} + i\sin\frac{\pi}{4} = \boxed{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\,i} \]

(b) \(e^{i\pi/2}\)

\[ e^{i\pi/2} = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2} = 0 + i = \boxed{i} \]

(c) \(e^{i\pi}\)

\[ e^{i\pi} = \cos\pi + i\sin\pi = -1 + 0i = \boxed{-1} \]

This corresponds to the famous Euler's identity \(e^{i\pi} + 1 = 0\).

(d) \(e^{-i\pi/3}\)

\[ e^{-i\pi/3} = \cos\frac{\pi}{3} - i\sin\frac{\pi}{3} = \boxed{\frac{1}{2} - \frac{\sqrt{3}}{2}\,i} \]

Verification: The absolute value of (a) \(= \sqrt{1/2 + 1/2} = 1\). This is consistent with \(|e^{i\theta}| = 1\). The others similarly have absolute value 1. ✓

B-9. Polar Form and Rewriting with Euler's Formula

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(a) \(z = 1 + i\)

\[ r = \sqrt{1+1} = \sqrt{2}, \quad \theta = \frac{\pi}{4} \quad (\text{first quadrant}) \]
\[ \boxed{z = \sqrt{2}\,e^{i\pi/4}} \]

(b) \(z = -\sqrt{3} + i\)

\[ r = \sqrt{3 + 1} = 2 \]

\(a = -\sqrt{3},\; b = 1\) (second quadrant). \(\tan\theta = \frac{1}{-\sqrt{3}}\), and since it is in the second quadrant, \(\theta = \frac{5\pi}{6}\).

\[ \boxed{z = 2\,e^{i \cdot 5\pi/6}} \]

Verification: \(2(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}) = 2(-\frac{\sqrt{3}}{2} + i\frac{1}{2}) = -\sqrt{3} + i\). ✓

(c) \(z = -5i\)

\[ r = 5, \quad \theta = -\frac{\pi}{2} \quad (\text{negative imaginary axis}) \]
\[ \boxed{z = 5\,e^{-i\pi/2}} \]

Verification: \(5(\cos(-\pi/2) + i\sin(-\pi/2)) = 5(0 - i) = -5i\). ✓

B-10. Calculating \(|e^{i\theta}|\)

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Strategy: Use the relation \(|z|^2 = zz^*\) from Eq. (A.15).

Let \(z = e^{i\theta}\), then \(z^* = e^{-i\theta}\) (from Euler's formula, the complex conjugate of \(e^{i\theta} = \cos\theta + i\sin\theta\) is \(\cos\theta - i\sin\theta = e^{-i\theta}\)).

\[ |e^{i\theta}|^2 = e^{i\theta} \cdot e^{-i\theta} = e^{i\theta - i\theta} = e^0 = 1 \]

Since \(|e^{i\theta}|\) is a non-negative real number,

\[ \boxed{|e^{i\theta}| = 1} \]

This holds for any real number \(\theta\). Geometrically, this means that \(e^{i\theta}\) always lies on the unit circle. \(\blacksquare\)

Medium

M-1. General Proof of the Product Rule for Complex Conjugates

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Proof:

Let \(z_1 = a + bi\) and \(z_2 = c + di\) (where \(a, b, c, d\) are real numbers).

Calculation of the left-hand side:

From equation (A.5),

\[ z_1 z_2 = (ac - bd) + (ad + bc)i \]

From the definition of the complex conjugate (equation (A.12)),

\[ (z_1 z_2)^* = (ac - bd) - (ad + bc)i \tag{*} \]

Calculation of the right-hand side:

\[ z_1^* = a - bi, \quad z_2^* = c - di \]
\[ z_1^* z_2^* = (a - bi)(c - di) = ac - adi - bci + bdi^2 \]
\[ = ac - adi - bci - bd = (ac - bd) - (ad + bc)i \tag{**} \]

Comparing \((*)\) and \((**)\),

\[ (z_1 z_2)^* = (ac - bd) - (ad + bc)i = z_1^* z_2^* \]

Therefore \((z_1 z_2)^* = z_1^* z_2^*\) holds. \(\blacksquare\)

Verification: This was confirmed concretely in D6(a) for the case \(z_1 = 2+i\), \(z_2 = 1-3i\). It is consistent with the general proof. ✓

M-2. Derivation of de Moivre's Theorem

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Proof of de Moivre's Theorem

Strategy: Using Euler's formula \(e^{i\theta} = \cos\theta + i\sin\theta\), rewrite the left-hand side in exponential form.

\[ (\cos\theta + i\sin\theta)^n = (e^{i\theta})^n \]

By the exponent rule \((e^a)^n = e^{na}\),

\[ (e^{i\theta})^n = e^{in\theta} \]

Applying Euler's formula again,

\[ e^{in\theta} = \cos(n\theta) + i\sin(n\theta) \]

Combining the above,

\[ \boxed{(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)} \]

holds for any integer \(n\). \(\blacksquare\)

The Case \(n = 2\): Derivation of the Double-Angle Formulas

Expanding the left-hand side:

\[ (\cos\theta + i\sin\theta)^2 = \cos^2\theta + 2i\cos\theta\sin\theta + i^2\sin^2\theta \]
\[ = (\cos^2\theta - \sin^2\theta) + 2i\cos\theta\sin\theta \]

Right-hand side (de Moivre's theorem):

\[ \cos 2\theta + i\sin 2\theta \]

Comparing real parts:

\[ \boxed{\cos 2\theta = \cos^2\theta - \sin^2\theta} \]

Comparing imaginary parts:

\[ \boxed{\sin 2\theta = 2\cos\theta\sin\theta} \]

These are precisely the double-angle formulas for trigonometric functions.

Verification: Check with \(\theta = \pi/4\). \(\cos(\pi/2) = 0\), \(\cos^2(\pi/4) - \sin^2(\pi/4) = 1/2 - 1/2 = 0\). ✓ \(\sin(\pi/2) = 1\), \(2\cos(\pi/4)\sin(\pi/4) = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = 1\). ✓

M-3. Derivation of the Division Formula in Polar Form

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Strategy: Use the laws of exponents to simplify the fraction.

\[ \frac{z_1}{z_2} = \frac{r_1 e^{i\theta_1}}{r_2 e^{i\theta_2}} \]

First, compute \(\frac{e^{i\theta_1}}{e^{i\theta_2}}\). Multiplying the numerator and denominator by \(e^{-i\theta_2}\),

\[ \frac{e^{i\theta_1}}{e^{i\theta_2}} = e^{i\theta_1} \cdot e^{-i\theta_2} = e^{i(\theta_1 - \theta_2)} \]

Here we used \(e^{i\theta_2} \cdot e^{-i\theta_2} = e^0 = 1\).

Therefore,

\[ \boxed{\frac{z_1}{z_2} = \frac{r_1}{r_2}\,e^{i(\theta_1 - \theta_2)}} \]

\(\blacksquare\)

Interpretation: From this result, for the quotient \(z_1/z_2\):

  • The modulus is \(\dfrac{r_1}{r_2} = \dfrac{|z_1|}{|z_2|}\) (division of moduli)
  • The argument is \(\theta_1 - \theta_2 = \arg(z_1) - \arg(z_2)\) (subtraction of arguments)

In other words, division of complex numbers decomposes into division of moduli and subtraction of arguments. This is the natural inverse operation of multiplication being "the product of moduli and the sum of arguments" (Eq. (A.11)).

Verification: When \(z_1 = z_2\), we have \(z_1/z_2 = 1\). From the formula: \(\frac{r_1}{r_1} e^{i \cdot 0} = 1 \cdot 1 = 1\). ✓

M-4. Exponential Representations of \(\cos\theta\) and \(\sin\theta\)

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Strategy: Solve Euler's formula and its complex conjugate as a system of simultaneous equations.

Euler's formula:

\[ e^{i\theta} = \cos\theta + i\sin\theta \tag{1} \]

Replace \(\theta\) with \(-\theta\) (or take the complex conjugate of equation (1)):

\[ e^{-i\theta} = \cos\theta - i\sin\theta \tag{2} \]

Derivation of \(\cos\theta\)

Equation (1) + Equation (2):

\[ e^{i\theta} + e^{-i\theta} = 2\cos\theta \]
\[ \boxed{\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}} \]

Derivation of \(\sin\theta\)

Equation (1) − Equation (2):

\[ e^{i\theta} - e^{-i\theta} = 2i\sin\theta \]
\[ \boxed{\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}} \]

Comparison with Equations (A.13) and (A.14)

Equation (A.13) states that \(\operatorname{Re}(z) = \frac{z + z^*}{2}\). Setting \(z = e^{i\theta}\) gives \(z^* = e^{-i\theta}\), so

\[ \operatorname{Re}(e^{i\theta}) = \frac{e^{i\theta} + e^{-i\theta}}{2} = \cos\theta \]

This is consistent with the fact that the real part of \(e^{i\theta} = \cos\theta + i\sin\theta\) is \(\cos\theta\).

Similarly, Equation (A.14) states that \(\operatorname{Im}(z) = \frac{z - z^*}{2i}\), and

\[ \operatorname{Im}(e^{i\theta}) = \frac{e^{i\theta} - e^{-i\theta}}{2i} = \sin\theta \]

This is also consistent with the fact that the imaginary part of \(e^{i\theta}\) is \(\sin\theta\).

From the above, we have confirmed that the exponential representations of \(\cos\theta\) and \(\sin\theta\) are nothing other than special cases (\(z = e^{i\theta}\), \(z^* = e^{-i\theta}\)) of the formulas for extracting real and imaginary parts using complex conjugates (Equations (A.13) and (A.14)). \(\blacksquare\)

Verification: Substitute \(\theta = 0\). \(\cos 0 = \frac{e^0 + e^0}{2} = \frac{2}{2} = 1\). ✓ \(\sin 0 = \frac{e^0 - e^0}{2i} = 0\). ✓

M-5. Conditions for a Complex Number to Be Real

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\(z\) is real \(\iff\) \(z = z^*\)

Proof:

Let \(z = a + bi\) (where \(a, b\) are real numbers).

(\(\Rightarrow\)) If \(z\) is real, then \(b = 0\), so \(z = a\) and \(z^* = a\). Therefore \(z = z^*\).

(\(\Leftarrow\)) If \(z = z^*\), then \(a + bi = a - bi\). Comparing the imaginary parts of both sides gives \(b = -b\), i.e., \(2b = 0\), so \(b = 0\). Therefore \(z = a\) is real. \(\blacksquare\)

\(z\) is purely imaginary (\(z \neq 0\)) \(\iff\) \(z = -z^*\) (and \(z \neq 0\))

Proof:

(\(\Rightarrow\)) If \(z\) is purely imaginary, then \(a = 0\) and \(b \neq 0\), so \(z = bi\) and \(z^* = -bi\). Therefore \(-z^* = bi = z\).

(\(\Leftarrow\)) If \(z = -z^*\), then \(a + bi = -(a - bi) = -a + bi\). Comparing the real parts of both sides gives \(a = -a\), i.e., \(2a = 0\), so \(a = 0\). From the condition \(z \neq 0\), we have \(b \neq 0\). Therefore \(z = bi\) is purely imaginary. \(\blacksquare\)

Verification (concrete examples): - \(z = 5\) (real): \(z^* = 5 = z\). ✓ - \(z = 3i\) (purely imaginary): \(z^* = -3i\), \(-z^* = 3i = z\). ✓ - \(z = 1 + i\) (neither): \(z^* = 1 - i \neq z\) and \(-z^* = -1 + i \neq z\). ✓

Advanced

A-1. The \(n\)-th Roots of a Complex Number and the Regular \(n\)-gon

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(a) The \(n\) solutions of \(z^n = 1\)

Strategy: Let \(z = re^{i\theta}\) and separate the conditions on the modulus and argument.

\[ z^n = r^n e^{in\theta} = 1 = 1 \cdot e^{i \cdot 0} \]

Comparing moduli: \(r^n = 1\). Since \(r > 0\), we have \(r = 1\).

Comparing arguments: \(n\theta = 0 + 2\pi k\) (\(k\) is an integer). Here the argument has an ambiguity of integer multiples of \(2\pi\).

\[ \theta = \frac{2\pi k}{n} \]

For \(k = 0, 1, 2, \ldots, n-1\), \(\theta\) takes distinct values \(0, \frac{2\pi}{n}, \frac{4\pi}{n}, \ldots, \frac{2\pi(n-1)}{n}\). When \(k = n\), \(\theta = 2\pi\) represents the same point as \(\theta = 0\), so no new solutions are obtained.

Therefore, the \(n\) solutions of \(z^n = 1\) are

\[ \boxed{z_k = e^{2\pi i k/n} \quad (k = 0, 1, 2, \ldots, n-1)} \]

\(\blacksquare\)

(b) Vertices of a regular \(n\)-gon

All \(n\) solutions \(z_k = e^{2\pi i k/n}\) satisfy \(|z_k| = 1\), so they are points on the unit circle.

The difference in argument between adjacent solutions \(z_k\) and \(z_{k+1}\) is

\[ \frac{2\pi(k+1)}{n} - \frac{2\pi k}{n} = \frac{2\pi}{n} \]

which is constant. That is, the \(n\) points are arranged at equal intervals on the unit circle.

A polygon inscribed in the unit circle with \(n\) vertices equally spaced is nothing other than a regular \(n\)-gon. The first vertex \(z_0 = 1\) is at the point \((1, 0)\) on the real axis, and the remaining vertices are located at positions rotated counterclockwise by \(2\pi/n\) from there. \(\blacksquare\)

(c) The sum of the \(n\)-th roots of unity is 0

Strategy: Use the geometric series formula.

Let \(\omega = e^{2\pi i/n}\), so that \(z_k = \omega^k\).

\[ \sum_{k=0}^{n-1} z_k = \sum_{k=0}^{n-1} \omega^k \]

This is a geometric series with first term \(1\), common ratio \(\omega\), and \(n\) terms, so (when \(\omega \neq 1\), i.e., \(n \geq 2\)),

\[ \sum_{k=0}^{n-1} \omega^k = \frac{1 - \omega^n}{1 - \omega} \]

Since \(\omega^n = (e^{2\pi i/n})^n = e^{2\pi i} = \cos 2\pi + i\sin 2\pi = 1\), we have

\[ \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = \frac{0}{1 - \omega} = 0 \]

When \(n = 1\), there is only \(z_0 = 1\) and the sum is \(1 \neq 0\), but we normally consider \(n \geq 2\).

\[ \boxed{\sum_{k=0}^{n-1} z_k = 0} \]

\(\blacksquare\)

Verification (geometric interpretation): By symmetry, the sum of position vectors to the vertices of a regular \(n\)-gon points to the origin. This corresponds to the centroid being at the origin. ✓

(d) The case \(n = 4\)

\[ z_k = e^{2\pi i k/4} = e^{i\pi k/2} \quad (k = 0, 1, 2, 3) \]
  • \(k = 0\): \(e^{0} = 1\)
  • \(k = 1\): \(e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2) = i\)
  • \(k = 2\): \(e^{i\pi} = \cos\pi + i\sin\pi = -1\)
  • \(k = 3\): \(e^{i3\pi/2} = \cos(3\pi/2) + i\sin(3\pi/2) = -i\)
\[ \boxed{\{z_0, z_1, z_2, z_3\} = \{1,\; i,\; -1,\; -i\}} \]

Verification: Sum \(= 1 + i + (-1) + (-i) = 0\). ✓ Also confirming \(z_k^4 = 1\) for each \(z_k\): \(i^4 = (i^2)^2 = (-1)^2 = 1\), \((-1)^4 = 1\), \((-i)^4 = ((-i)^2)^2 = (-1)^2 = 1\). ✓

A-2. Bridge to Quantum Mechanics: Interference of Complex Amplitudes

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(a) Expansion of \(P = |\phi_1 + \phi_2|^2\)

Approach: Expand using \(|\phi|^2 = \phi\phi^*\).

\[ P = |\phi_1 + \phi_2|^2 = (\phi_1 + \phi_2)(\phi_1 + \phi_2)^* = (\phi_1 + \phi_2)(\phi_1^* + \phi_2^*) \]

Expanding,

\[ P = \phi_1\phi_1^* + \phi_1\phi_2^* + \phi_2\phi_1^* + \phi_2\phi_2^* \]
\[ = |\phi_1|^2 + |\phi_2|^2 + \phi_1\phi_2^* + (\phi_1\phi_2^*)^* \]

Now substitute \(\phi_1 = r_1 e^{i\alpha}\), \(\phi_2 = r_2 e^{i\beta}\).

\[ |\phi_1|^2 = r_1^2, \quad |\phi_2|^2 = r_2^2 \]
\[ \phi_1\phi_2^* = r_1 e^{i\alpha} \cdot r_2 e^{-i\beta} = r_1 r_2 e^{i(\alpha - \beta)} \]
\[ \phi_2\phi_1^* = r_1 r_2 e^{-i(\alpha - \beta)} = (\phi_1\phi_2^*)^* \]

The sum of the cross terms is,

\[ \phi_1\phi_2^* + \phi_2\phi_1^* = r_1 r_2 \left[e^{i(\alpha-\beta)} + e^{-i(\alpha-\beta)}\right] \]

Using the exponential representation of \(\cos\) derived in S4, \(\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}\),

\[ e^{i(\alpha-\beta)} + e^{-i(\alpha-\beta)} = 2\cos(\alpha - \beta) \]

Therefore,

\[ \boxed{P = r_1^2 + r_2^2 + 2r_1 r_2 \cos(\alpha - \beta)} \]

\(\blacksquare\)

Verification: When \(\phi_2 = 0\) (no path 2), \(r_2 = 0\) so \(P = r_1^2 = |\phi_1|^2\). ✓ Also when \(\alpha = \beta\) (in phase), \(P = r_1^2 + r_2^2 + 2r_1 r_2 = (r_1 + r_2)^2\). This corresponds to the amplitudes adding together. ✓

(b) Comparison between real and complex amplitudes

Real amplitude case (where \(\alpha, \beta\) are only \(0\) or \(\pi\)):

The possible values of \(\alpha - \beta\) are \(0, \pi, -\pi\).

  • When \(\alpha - \beta = 0\): \(\cos(\alpha - \beta) = \cos 0 = 1\)
  • When \(\alpha - \beta = \pm\pi\): \(\cos(\alpha - \beta) = \cos\pi = -1\)

Therefore the interference term \(2r_1 r_2 \cos(\alpha - \beta)\) takes only two values: \(+2r_1 r_2\) or \(-2r_1 r_2\).

\[ \text{Real case: interference term} \in \{-2r_1 r_2,\; +2r_1 r_2\} \]

Complex amplitude case (where \(\alpha - \beta\) varies continuously):

The phase difference \(\delta = \alpha - \beta\) can vary continuously from \(0\) to \(2\pi\), and \(\cos\delta\) takes all values continuously from \(-1\) to \(+1\). Therefore the interference term varies continuously over the range

\[ -2r_1 r_2 \leq 2r_1 r_2 \cos(\alpha - \beta) \leq +2r_1 r_2 \]

Why complex numbers are essentially necessary: With real amplitudes, interference is limited to two choices—"completely constructive" or "completely destructive"—and intermediate interference cannot be expressed. On the other hand, with complex amplitudes, the phase difference becomes a continuous parameter, enabling the description of all degrees of interference, including partial constructive and destructive interference. To correctly describe the interference patterns observed in nature (such as the continuous variation of bright and dark fringes in the double-slit experiment), it is essential that amplitudes be complex numbers.

(c) Complete destructive interference

When \(r_1 = r_2 = r\), the result from (a) becomes

\[ P = r^2 + r^2 + 2r^2\cos(\alpha - \beta) = 2r^2(1 + \cos(\alpha - \beta)) \]

Substituting \(\alpha - \beta = \pi\),

\[ P = 2r^2(1 + \cos\pi) = 2r^2(1 - 1) = \boxed{0} \]

When the amplitudes of two paths are equal and the phase difference is exactly \(\pi\), the detection probability becomes exactly zero. This is destructive interference.

Contrast with classical intuition:

In classical probability theory, one simply adds the probabilities of passing through each path as \(P_1 = r^2\) and \(P_2 = r^2\), giving

\[ P_{\text{classical}} = P_1 + P_2 = 2r^2 > 0 \]

so the probability can never be zero. However, in quantum mechanics, what we add are not probabilities but probability amplitudes (complex numbers), and cancellation can occur at the amplitude level, resulting in zero probability. This is a phenomenon that contradicts the classical intuition that "the particle must pass through one path or the other, yet is never detected," and it is an essential feature of quantum mechanics.

Verification: When \(\alpha - \beta = 0\) (constructive interference), \(P = 2r^2(1+1) = 4r^2 = (2r)^2\). The amplitude doubles to \(2r\), and the probability quadruples. This is twice the classical value of \(2r^2\), clearly showing the effect of interference. ✓