S.M. Yusuf Mathematical Methods Chapter 1 Complex Numbers & De Moivre Theorem Solved Exercises PDF Guide | BS Math & BSc Past Papers (PU, UOS, KU, QAU, FBISE)

Theoretical Foundations & Core Theorems

Complex analysis forms the bedrock of modern pure and applied mathematics, physics, and engineering. Chapter 1 of Mathematical Methods by S.M. Yusuf introduces the formal system of complex numbers, expanding the field of real numbers $\mathbb{R}$ to the algebraically closed field of complex numbers $\mathbb{C}$.

1. The Field of Complex Numbers ($\mathbb{C}$)

A complex number $z$ is defined as an ordered pair of real numbers $(x, y) \in \mathbb{R}^2$, expressed in standard form as:

$$z = x + iy$$

where $x = \text{Re}(z)$ represents the real part, $y = \text{Im}(z)$ represents the imaginary part, and $i = \sqrt{-1}$ is the imaginary unit satisfying $i^2 = -1$. The complex conjugate of $z$ is given by $\bar{z} = x - iy$, yielding the crucial property that $z\bar{z} = x^2 + y^2 = |z|^2$, where $|z|$ is the modulus (or absolute value) of $z$.

2. Polar and Exponential Representations

By mapping $z = x + iy$ to a point $(x, y)$ in the Argand plane, we transform Cartesian coordinates to polar coordinates $(r, \theta)$:

$$x = r\cos\theta, \quad y = r\sin\theta$$

where $r = |z| = \sqrt{x^2 + y^2} \geq 0$ is the magnitude, and $\theta = \arg(z)$ is the argument of $z$. The principal argument, denoted by $\text{Arg}(z) = \Theta$, is uniquely restricted to the interval $-\pi < \Theta \leq \pi$. Using Euler's Formula ($e^{i\theta} = \cos\theta + i\sin\theta$), any complex number can be expressed exponentially as:

$$z = r(\cos\theta + i\sin\theta) = r e^{i\theta}$$

3. De Moivre's Theorem

De Moivre's Theorem is a cornerstone of complex algebra, bridging complex exponentials and trigonometry.

Theorem Statement:

1. For any integer $n \in \mathbb{Z}$: $$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$
2. For any rational number $n = \frac{p}{q}$ (where $q > 0$ and $\gcd(p, q) = 1$): One of the values of $(\cos\theta + i\sin\theta)^{p/q}$ is $\cos\left(\frac{p\theta}{q}\right) + i\sin\left(\frac{p\theta}{q}\right)$. The full set of $q$ distinct values is generated by: $$\left(\cos\theta + i\sin\theta\right)^{p/q} = \cos\left(\frac{p(\theta + 2k\pi)}{q}\right) + i\sin\left(\frac{p(\theta + 2k\pi)}{q}\right), \quad k = 0, 1, 2, \dots, q-1$$

4. Roots of Complex Numbers

To find the $n$-th roots of a complex number $z = r(\cos\theta + i\sin\theta)$, we solve $w^n = z$. Applying De Moivre's Theorem yields $n$ distinct roots spaced evenly by $\frac{2\pi}{n}$ radians on a circle of radius $r^{1/n}$ in the complex plane:

$$w_k = r^{1/n} \left[ \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right], \quad k = 0, 1, 2, \dots, n-1$$

5. Expansions of $\cos(n\theta)$, $\sin(n\theta)$, $\cos^n\theta$, and $\sin^n\theta$

Let $x = e^{i\theta} = \cos\theta + i\sin\theta$. Then $\frac{1}{x} = e^{-i\theta} = \cos\theta - i\sin\theta$. From these relations, we derive:

$$x + \frac{1}{x} = 2\cos\theta \quad \text{and} \quad x - \frac{1}{x} = 2i\sin\theta$$ $$x^n + \frac{1}{x^n} = 2\cos(n\theta) \quad \text{and} \quad x^n - \frac{1}{x^n} = 2i\sin(n\theta)$$

Expanding $(x + x^{-1})^n$ or $(x - x^{-1})^n$ via the Binomial Theorem allows us to express powers of trigonometric functions as sums of sines or cosines of multiple angles.

---

Formula Summary & Quick Reference

Key Concept	Formula / Identity	Academic Application
Modulus & Conjugate	$|z| = \sqrt{x^2+y^2}$, $\bar{z} = x - iy$, $z\bar{z} = |z|^2$	Simplifying complex fractions, proving distance properties.
Principal Argument ($\Theta$)	$\Theta = \begin{cases} \alpha & \text{Quadrant I } (x>0, y>0) \\ \pi - \alpha & \text{Quadrant II } (x<0, y>0) \\ -\pi + \alpha & \text{Quadrant III } (x<0, y<0) \\ -\alpha & \text{Quadrant IV } (x>0, y<0) \end{cases}$ where $\alpha = \tan^{-1}\left|\frac{y}{x}\right|$	Converting Cartesian form $x+iy$ to standard polar form $r(\cos\Theta + i\sin\Theta)$.
De Moivre's Theorem	$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$	Simplifying expressions, evaluating high integral powers of complex numbers.
$n$-th Roots of $z$	$z_k = r^{1/n} \text{cis}\left(\frac{\theta + 2k\pi}{n}\right), \quad k = 0, 1, \dots, n-1$	Solving polynomial equations of the form $z^n - w = 0$.
Multiple Angle Expansions	$x^n + x^{-n} = 2\cos(n\theta)$, $x^n - x^{-n} = 2i\sin(n\theta)$	Expressing $\cos^n\theta$ or $\sin^n\theta$ in linear combinations of $\cos(k\theta)$ or $\sin(k\theta)$.

---

Step-by-Step Solved Board Exam Questions

Question 1 (PU / FBISE Past Paper)

Problem: Prove that:

$$\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n = \cos\left(n\left(\frac{\pi}{2} - \theta\right)\right) + i\sin\left(n\left(\frac{\pi}{2} - \theta\right)\right)$$

Solution:

Step 1: Express the numerator and denominator in polar form.
Recall the identity $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ and $\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)$. Let $\phi = \frac{\pi}{2} - \theta$.
Then the numerator $N$ becomes:

$$N = 1 + \cos\phi + i\sin\phi$$

Using half-angle trigonometric formulas $1 + \cos\phi = 2\cos^2\left(\frac{\phi}{2}\right)$ and $\sin\phi = 2\sin\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)$:

$$N = 2\cos^2\left(\frac{\phi}{2}\right) + i \cdot 2\sin\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right) = 2\cos\left(\frac{\phi}{2}\right) \left[ \cos\left(\frac{\phi}{2}\right) + i\sin\left(\frac{\phi}{2}\right) \right]$$

Similarly, the denominator $D$ is:

$$D = 1 + \cos\phi - i\sin\phi = 2\cos\left(\frac{\phi}{2}\right) \left[ \cos\left(\frac{\phi}{2}\right) - i\sin\left(\frac{\phi}{2}\right) \right]$$

Step 2: Simplify the fraction $\frac{N}{D}$.

$$\frac{N}{D} = \frac{2\cos\left(\frac{\phi}{2}\right) \left[ \cos\left(\frac{\phi}{2}\right) + i\sin\left(\frac{\phi}{2}\right) \right]}{2\cos\left(\frac{\phi}{2}\right) \left[ \cos\left(\frac{\phi}{2}\right) - i\sin\left(\frac{\phi}{2}\right) \right]} = \frac{\cos\left(\frac{\phi}{2}\right) + i\sin\left(\frac{\phi}{2}\right)}{\cos\left(\frac{\phi}{2}\right) - i\sin\left(\frac{\phi}{2}\right)}$$

Using negative arguments, $\cos\left(\frac{\phi}{2}\right) - i\sin\left(\frac{\phi}{2}\right) = \cos\left(-\frac{\phi}{2}\right) + i\sin\left(-\frac{\phi}{2}\right) = e^{-i\phi/2}$. Thus:

$$\frac{N}{D} = \frac{e^{i\phi/2}}{e^{-i\phi/2}} = e^{i\phi} = \cos\phi + i\sin\phi$$

Step 3: Apply the power $n$ using De Moivre's Theorem.

$$\left(\frac{N}{D}\right)^n = (\cos\phi + i\sin\phi)^n = \cos(n\phi) + i\sin(n\phi)$$

Step 4: Substitute $\phi = \frac{\pi}{2} - \theta$ back into the expression.

$$\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n = \cos\left(n\left(\frac{\pi}{2} - \theta\right)\right) + i\sin\left(n\left(\frac{\pi}{2} - \theta\right)\right)$$

Q.E.D. (Proved)

---

Question 2 (UOS / KU Past Paper)

Problem: Expand $\cos^6\theta$ in a series of cosines of multiples of $\theta$.

Solution:

Step 1: Set up the fundamental relation.
Let $x = \cos\theta + i\sin\theta = e^{i\theta}$. Then $\frac{1}{x} = \cos\theta - i\sin\theta$.
Addition gives:

$$x + \frac{1}{x} = 2\cos\theta \implies (2\cos\theta)^6 = \left(x + \frac{1}{x}\right)^6$$

Step 2: Expand using the Binomial Theorem.
The Binomial expansion formula is $(a+b)^6 = \sum_{k=0}^6 \binom{6}{k} a^{6-k} b^k$.

$$64\cos^6\theta = x^6 + \binom{6}{1}x^5\left(\frac{1}{x}\right) + \binom{6}{2}x^4\left(\frac{1}{x^2}\right) + \binom{6}{3}x^3\left(\frac{1}{x^3}\right) + \binom{6}{4}x^2\left(\frac{1}{x^4}\right) + \binom{6}{5}x\left(\frac{1}{x^5}\right) + \frac{1}{x^6}$$

Calculating coefficients: $\binom{6}{1}=6, \binom{6}{2}=15, \binom{6}{3}=20, \binom{6}{4}=15, \binom{6}{5}=6$.

$$64\cos^6\theta = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$$

Step 3: Group terms with symmetric powers.

$$64\cos^6\theta = \left(x^6 + \frac{1}{x^6}\right) + 6\left(x^4 + \frac{1}{x^4}\right) + 15\left(x^2 + \frac{1}{x^2}\right) + 20$$

Step 4: Substitute $x^k + \frac{1}{x^k} = 2\cos(k\theta)$.

$$64\cos^6\theta = 2\cos(6\theta) + 6(2\cos(4\theta)) + 15(2\cos(2\theta)) + 20$$ $$64\cos^6\theta = 2\cos(6\theta) + 12\cos(4\theta) + 30\cos(2\theta) + 20$$

Step 5: Divide by 64.

$$\cos^6\theta = \frac{1}{32} \left[ \cos(6\theta) + 6\cos(4\theta) + 15\cos(2\theta) + 10 \right]$$

---

Question 3 (QAU / PU Past Paper)

Problem: Solve the equation $z^7 + 1 = 0$ using De Moivre's Theorem and show that the roots form a geometric progression.

Solution:

Step 1: Rewrite the equation in polar form.

$$z^7 = -1$$

The number $-1$ in polar form has modulus $r = 1$ and principal argument $\theta = \pi$:

$$-1 = \cos\pi + i\sin\pi$$

Including the general period $2k\pi$ ($k \in \mathbb{Z}$):

$$z^7 = \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) = \cos((2k+1)\pi) + i\sin((2k+1)\pi)$$

Step 2: Apply De Moivre's Theorem for roots.

$$z_k = \cos\left(\frac{(2k+1)\pi}{7}\right) + i\sin\left(\frac{(2k+1)\pi}{7}\right), \quad k = 0, 1, 2, 3, 4, 5, 6$$

Step 3: List the 7 distinct roots explicit in exponential form.
Let $\alpha = e^{i\pi/7} = \cos\left(\frac{\pi}{7}\right) + i\sin\left(\frac{\pi}{7}\right)$.

• $k = 0 \implies z_0 = e^{i\pi/7} = \alpha$
• $k = 1 \implies z_1 = e^{i3\pi/7} = \alpha^3$
• $k = 2 \implies z_2 = e^{i5\pi/7} = \alpha^5$
• $k = 3 \implies z_3 = e^{i7\pi/7} = e^{i\pi} = -1 = \alpha^7$
• $k = 4 \implies z_4 = e^{i9\pi/7} = \alpha^9$
• $k = 5 \implies z_5 = e^{i11\pi/7} = \alpha^{11}$
• $k = 6 \implies z_6 = e^{i13\pi/7} = \alpha^{13}$

Step 4: Prove Geometric Progression.
The set of roots is $\{\alpha, \alpha^3, \alpha^5, \alpha^7, \alpha^9, \alpha^{11}, \alpha^{13}\}$.
The ratio between consecutive roots is:

$$\frac{z_{k+1}}{z_k} = \frac{\alpha^{2k+3}}{\alpha^{2k+1}} = \alpha^2 = e^{i2\pi/7}$$

Since the common ratio $r = \alpha^2$ is constant, the roots form a Geometric Progression (G.P.).

---

Question 4 (S.M. Yusuf Ex 1.3 / PU Long Question)

Problem: If $\cos\alpha + \cos\beta + \cos\gamma = 0$ and $\sin\alpha + \sin\beta + \sin\gamma = 0$, prove that:

$$\cos(3\alpha) + \cos(3\beta) + \cos(3\gamma) = 3\cos(\alpha + \beta + \gamma)$$ $$\sin(3\alpha) + \sin(3\beta) + \sin(3\gamma) = 3\sin(\alpha + \beta + \gamma)$$

Solution:

Step 1: Define complex variables.
Let $a = e^{i\alpha} = \cos\alpha + i\sin\alpha$, $b = e^{i\beta} = \cos\beta + i\sin\beta$, and $c = e^{i\gamma} = \cos\gamma + i\sin\gamma$.

Step 2: Sum the variables.

$$a + b + c = (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + \sin\beta + \sin\gamma)$$

Given that both real and imaginary sums equal $0$:

$$a + b + c = 0 + i(0) = 0$$

Step 3: Apply the algebraic identity for sum of cubes.
Recall the classical algebraic identity: If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Step 4: Evaluate $a^3, b^3, c^3$ and $abc$ using De Moivre's Theorem.

• $a^3 = (e^{i\alpha})^3 = e^{i3\alpha} = \cos(3\alpha) + i\sin(3\alpha)$
• $b^3 = (e^{i\beta})^3 = e^{i3\beta} = \cos(3\beta) + i\sin(3\beta)$
• $c^3 = (e^{i\gamma})^3 = e^{i3\gamma} = \cos(3\gamma) + i\sin(3\gamma)$

For the right-hand side:

$$abc = e^{i\alpha} \cdot e^{i\beta} \cdot e^{i\gamma} = e^{i(\alpha + \beta + \gamma)} = \cos(\alpha + \beta + \gamma) + i\sin(\alpha + \beta + \gamma)$$

Step 5: Substitute back into $a^3 + b^3 + c^3 = 3abc$.

$$[\cos(3\alpha) + \cos(3\beta) + \cos(3\gamma)] + i[\sin(3\alpha) + \sin(3\beta) + \sin(3\gamma)] = 3[\cos(\alpha + \beta + \gamma) + i\sin(\alpha + \beta + \gamma)]$$

Step 6: Equate real and imaginary parts.
Equating Real parts:

$$\cos(3\alpha) + \cos(3\beta) + \cos(3\gamma) = 3\cos(\alpha + \beta + \gamma)$$

Equating Imaginary parts:

$$\sin(3\alpha) + \sin(3\beta) + \sin(3\gamma) = 3\sin(\alpha + \beta + \gamma)$$

Q.E.D. (Proved)

---

Interactive Practice Quiz (Clickable MCQs)

---

Frequently Asked Questions (FAQs)

1. How do I reliably determine the Principal Argument $\text{Arg}(z)$ without making quadrant sign errors?

To avoid common sign errors in university board examinations (PU, UOS, FBISE):

1. First calculate the acute reference angle $\alpha = \tan^{-1}\left|\frac{y}{x}\right|$, ignoring signs of $x$ and $y$.
2. Identify the quadrant of $z = x + iy$ based on the signs of $x$ and $y$:
   • Quadrant I ($x>0, y>0$): $\Theta = \alpha$
   • Quadrant II ($x<0, y>0$): $\Theta = \pi - \alpha$
   • Quadrant III ($x<0, y<0$): $\Theta = -\pi + \alpha$ (or $\alpha - \pi$)
   • Quadrant IV ($x>0, y<0$): $\Theta = -\alpha$
3. Verify that your final angle lies strictly in the principal range $(-\pi, \pi]$. Do not write arguments as $240^\circ$ ($\frac{4\pi}{3}$); instead, convert to negative principal value $-\frac{2\pi}{3}$.

2. Why does evaluating rational powers $z^{p/q}$ require adding $2k\pi$ before applying De Moivre's Theorem?

When power $n = \frac{p}{q}$ is fractional/rational, the expression $z^{p/q}$ represents a multi-valued function with exactly $q$ distinct complex roots. If you apply De Moivre's Theorem directly to $r^{p/q}(\cos\theta + i\sin\theta)^{p/q}$ without adding $2k\pi$, you obtain only one principal root. Adding $2k\pi$ to the argument ($\theta + 2k\pi$) and iterating $k = 0, 1, 2, \dots, q-1$ ensures all $q$ unique roots distributed symmetrically around the origin are correctly generated.

3. How are marks distributed for De Moivre's proof questions in Pakistani University exams?

In 10-mark or 8-mark long questions (e.g., Ex 1.3/1.4 of S.M. Yusuf):

• Formulation & Conversion (20%): Explicitly setting up complex identities $x = e^{i\theta}$, $x + x^{-1} = 2\cos\theta$, or expressing terms in $r e^{i\theta}$ form.
• Algebraic Manipulation & Binomial Expansion (40%): Writing out step-by-step expansions, evaluating combinations $\binom{n}{k}$, and grouping symmetric power terms like $(x^k + x^{-k})$.
• Application of De Moivre's Theorem (20%): Substituting $x^k + x^{-k} = 2\cos(k\theta)$ or applying $(\text{cis}\theta)^n = \text{cis}(n\theta)$.
• Final Conclusion & Equating Parts (20%): Equating real/imaginary components cleanly and stating Q.E.D. Skipping steps or writing direct answers without showing the binomial expansion will lead to marks deduction.