S.M. Yusuf Mathematical Methods Chapter 1 Complex Numbers & De Moivre Theorem Solved Exercises PDF Guide | BS Math PU UOS KU QAU & Board Exams

Theoretical Foundations & Core Theorems

Complex analysis forms the bedrock of modern pure and applied mathematics. In the curriculum prescribed by Pakistani universities (PU, UOS, KU, QAU) following the textbook Mathematical Methods by S.M. Yusuf, Chapter 1 bridges elementary algebraic structures with advanced analytical methods. A complex number $z \in \mathbb{C}$ is formally defined as an ordered pair of real numbers $(x, y)$, written in standard form as:

$$z = x + i y \quad \text{where } x, y \in \mathbb{R} \text{ and } i = \sqrt{-1}$$

1. The Argand Plane and Polar Representation

Geometrically, complex numbers map onto a two-dimensional real vector space known as the Argand plane. The transformation from Cartesian coordinates $(x, y)$ to Polar coordinates $(r, \theta)$ is established via the mappings $x = r \cos\theta$ and $y = r \sin\theta$. Consequently, any complex number $z$ can be expressed as:

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

where:

• Modulus: $|z| = r = \sqrt{x^2 + y^2}$, representing the Euclidean distance from the origin $(0,0)$ to the point $(x,y)$.
• Principal Argument: $\text{Arg}(z) = \theta \in (-\pi, \pi]$, which represents the oriented angle between the positive real axis and the position vector of $z$. The general argument is expressed as $\arg(z) = \text{Arg}(z) + 2k\pi$ for $k \in \mathbb{Z}$.

2. Euler's Formula and Exponential Form

Euler's identity forms the connective tissue between complex exponentials and trigonometric functions. Formally defined via Maclaurin series expansions:

$$e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} = \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots \right) + i \left( \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots \right) = \cos\theta + i \sin\theta$$

3. De Moivre's Theorem

Theorem Statement: For any real number $n \in \mathbb{R}$ and any angle $\theta \in \mathbb{R}$:

$$(\cos\theta + i \sin\theta)^n = \cos(n\theta) + i \sin(n\theta)$$

Proof Case for Integer $n \in \mathbb{Z}^+$ (Mathematical Induction):

1. Base Step: For $n=1$, $(\cos\theta + i\sin\theta)^1 = \cos(1\cdot\theta) + i\sin(1\cdot\theta)$, which holds trivially.
2. Inductive Hypothesis: Assume true for $n = k$, i.e., $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$.
3. Inductive Step: For $n = k+1$: $$(\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k \cdot (\cos\theta + i\sin\theta) = [\cos(k\theta) + i\sin(k\theta)][\cos\theta + i\sin\theta]$$ Expanding the product: $$= \cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta + i[\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta]$$ Applying standard trigonometric addition formulas: $$= \cos((k+1)\theta) + i\sin((k+1)\theta)$$ Hence, by mathematical induction, the theorem holds for all positive integers. Extending to negative integers ($n = -m, m \in \mathbb{Z}^+$) and rational numbers ($n = p/q$) follows directly via field axioms.

4. Roots of Complex Numbers

To compute the $n$-th roots of a complex number $z = r e^{i\theta}$, we solve $w^n = z$. Applying De Moivre's Theorem yields $n$ distinct roots given by:

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

Geometrically, these $n$ roots form the vertices of a regular $n$-sided polygon inscribed within a circle of radius $r^{1/n}$ centered at the origin in the complex plane.

Formula Summary & Quick Reference

Key Concept	Mathematical Formula / Identity	Application Scope
Modulus-Argument Form	$z = r(\cos\theta + i\sin\theta)$, $r = \sqrt{x^2+y^2}$	Converting rectangular to polar form
Euler's Identity	$e^{i\theta} = \cos\theta + i\sin\theta$, $e^{i\pi} + 1 = 0$	Complex exponentials & analytical mechanics
De Moivre's Theorem	$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$	Powers of complex numbers & trig identities
$n$-th Roots Identity	$z_k = r^{1/n} \exp\left(i \frac{\theta + 2k\pi}{n}\right)$, $k=0,\dots,n-1$	Solving polynomial equations $z^n = a + ib$
Trigonometric Powers	$2\cos\theta = x + x^{-1}$, $2i\sin\theta = x - x^{-1}$ where $x = e^{i\theta}$	Expanding $\cos^n\theta$ and $\sin^n\theta$ series
Triangle Inequality	$||z_1| - |z_2|| \le |z_1 \pm z_2| \le |z_1| + |z_2|$	Bounding complex expressions & limits

Step-by-Step Solved Board Exam Questions

Question 1 (S.M. Yusuf Ch 1 Exercise 1.2, Past University Exam)

Problem Statement: Prove that:

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

Solution:

Step 1: Apply half-angle trigonometric identities to simplify $1 + \cos\theta$ and $\sin\theta$.

Recall that $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$ and $\sin\theta = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$.

Substitute these identities into the first expression $A = 1 + \cos\theta + i\sin\theta$:

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

Step 2: Raise expression $A$ to the $n$-th power and apply De Moivre's Theorem:

$$A^n = \left[ 2\cos\left(\frac{\theta}{2}\right) \right]^n \left[ \cos\left(\frac{\theta}{2}\right) + i\sin\left(\frac{\theta}{2}\right) \right]^n$$ $$A^n = 2^n \cos^n\left(\frac{\theta}{2}\right) \left[ \cos\left(\frac{n\theta}{2}\right) + i\sin\left(\frac{n\theta}{2}\right) \right] \quad \text{--- (Equation 1)}$$

Step 3: Similarly, simplify the second term $B = 1 + \cos\theta - i\sin\theta$ and raise it to the $n$-th power:

$$B = 2\cos\left(\frac{\theta}{2}\right) \left[ \cos\left(\frac{\theta}{2}\right) - i\sin\left(\frac{\theta}{2}\right) \right]$$ $$B^n = 2^n \cos^n\left(\frac{\theta}{2}\right) \left[ \cos\left(\frac{n\theta}{2}\right) - i\sin\left(\frac{n\theta}{2}\right) \right] \quad \text{--- (Equation 2)}$$

Step 4: Add Equation 1 and Equation 2:

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

Conclusion: Hence, the identity is completely proved. $\blacksquare$

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Question 2 (Classic Past Paper Problem: Tri-Angle Sums)

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

$$\cos(3\alpha) + \cos(3\beta) + \cos(3\gamma) = 3\cos(\alpha + \beta + \gamma)$$ $$\text{and } \sin(3\alpha) + \sin(3\beta) + \sin(3\gamma) = 3\sin(\alpha + \beta + \gamma)$$

Solution:

Step 1: Define three complex variables $a, b, c$ in Euler polar form:

$$a = e^{i\alpha} = \cos\alpha + i\sin\alpha$$ $$b = e^{i\beta} = \cos\beta + i\sin\beta$$ $$c = e^{i\gamma} = \cos\gamma + i\sin\gamma$$

Step 2: Sum the variables $a, b, c$:

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

Given that the real and imaginary parts both vanish:

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

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

Substitute the definitions of $a, b, c$ using De Moivre's Theorem:

Left-Hand Side (LHS):

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

Right-Hand Side (RHS):

$$3abc = 3 (e^{i\alpha} \cdot e^{i\beta} \cdot e^{i\gamma}) = 3 e^{i(\alpha + \beta + \gamma)}$$ $$3abc = 3 [\cos(\alpha + \beta + \gamma) + i\sin(\alpha + \beta + \gamma)]$$

Step 4: Equate real and imaginary components from LHS = RHS:

$$\text{Real Part: } \cos(3\alpha) + \cos(3\beta) + \cos(3\gamma) = 3\cos(\alpha + \beta + \gamma)$$ $$\text{Imaginary Part: } \sin(3\alpha) + \sin(3\beta) + \sin(3\gamma) = 3\sin(\alpha + \beta + \gamma)$$

Conclusion: Both required identities are proved simultaneously. $\blacksquare$

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Question 3 (Roots of Unity & Factorization)

Problem Statement: Solve the equation $z^7 + 1 = 0$ using De Moivre's Theorem, and express all solutions explicitly.

Solution:

Step 1: Rearrange the equation to isolate $z^7$:

$$z^7 = -1$$

Step 2: Express $-1$ in polar exponential form. The magnitude $r = 1$ and argument $\theta = \pi$. Including the general $2k\pi$ periodicity:

$$-1 = \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) = e^{i(2k+1)\pi}, \quad k \in \mathbb{Z}$$

Step 3: Take the $1/7$-th power on both sides and apply De Moivre's Theorem:

$$z_k = \left[ \cos((2k+1)\pi) + i\sin((2k+1)\pi) \right]^{1/7}$$ $$z_k = \cos\left(\frac{(2k+1)\pi}{7}\right) + i\sin\left(\frac{(2k+1)\pi}{7}\right) = \exp\left(i\frac{(2k+1)\pi}{7}\right)$$

Step 4: Evaluate $z_k$ for the seven distinct values of $k = 0, 1, 2, 3, 4, 5, 6$:

• For $k = 0$: $z_0 = \cos\left(\frac{\pi}{7}\right) + i\sin\left(\frac{\pi}{7}\right) = e^{i\pi/7}$
• For $k = 1$: $z_1 = \cos\left(\frac{3\pi}{7}\right) + i\sin\left(\frac{3\pi}{7}\right) = e^{i 3\pi/7}$
• For $k = 2$: $z_2 = \cos\left(\frac{5\pi}{7}\right) + i\sin\left(\frac{5\pi}{7}\right) = e^{i 5\pi/7}$
• For $k = 3$: $z_3 = \cos\left(\frac{7\pi}{7}\right) + i\sin\left(\frac{7\pi}{7}\right) = \cos(\pi) + i\sin(\pi) = -1$
• For $k = 4$: $z_4 = \cos\left(\frac{9\pi}{7}\right) + i\sin\left(\frac{9\pi}{7}\right) = e^{i 9\pi/7}$
• For $k = 5$: $z_5 = \cos\left(\frac{11\pi}{7}\right) + i\sin\left(\frac{11\pi}{7}\right) = e^{i 11\pi/7}$
• For $k = 6$: $z_6 = \cos\left(\frac{13\pi}{7}\right) + i\sin\left(\frac{13\pi}{7}\right) = e^{i 13\pi/7}$

Final Solution Set: $S.S. = \left\{ \exp\left(i\frac{(2k+1)\pi}{7}\right) \;\middle|\; k = 0, 1, 2, 3, 4, 5, 6 \right\}$. $\blacksquare$

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Question 4 (Trigonometric Power Expansion)

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

Solution:

Step 1: Let $x = e^{i\theta} = \cos\theta + i\sin\theta$. Then $x^{-1} = \frac{1}{x} = e^{-i\theta} = \cos\theta - i\sin\theta$.

By adding these definitions:

$$x + \frac{1}{x} = 2\cos\theta$$

Also, by De Moivre's Theorem, for any integer $k$:

$$x^k + \frac{1}{x^k} = 2\cos(k\theta)$$

Step 2: Raise $2\cos\theta$ to the 6th power:

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

Step 3: Expand using the Binomial Theorem $(a+b)^6 = \sum_{r=0}^{6} \binom{6}{r} a^{6-r} b^r$ with coefficients $[1, 6, 15, 20, 15, 6, 1]$:

$$64 \cos^6\theta = x^6 + 6 x^5 \left(\frac{1}{x}\right) + 15 x^4 \left(\frac{1}{x^2}\right) + 20 x^3 \left(\frac{1}{x^3}\right) + 15 x^2 \left(\frac{1}{x^4}\right) + 6 x \left(\frac{1}{x^5}\right) + \frac{1}{x^6}$$ $$64 \cos^6\theta = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}$$

Step 4: Group complementary power terms $(x^k + x^{-k})$ together:

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

Step 5: Substitute $x^k + \frac{1}{x^k} = 2\cos(k\theta)$ into the grouped expansion:

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

Divide the entire equation by 64:

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

Conclusion: The expansion of $\cos^6\theta$ in terms of multiple angles is derived completely. $\blacksquare$

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

1. How do I determine the correct quadrant for the Principal Argument $\text{Arg}(z)$?

To avoid common sign errors when computing $\theta = \text{Arg}(z)$ for $z = x + iy$:

1. First compute the acute reference angle $\alpha = \tan^{-1}\left|\frac{y}{x}\right|$.
2. Locate the quadrant based on the real algebraic signs of $x$ and $y$:
   • Quadrant I ($x>0, y>0$): $\text{Arg}(z) = \alpha$
   • Quadrant II ($x<0, y>0$): $\text{Arg}(z) = \pi - \alpha$
   • Quadrant III ($x<0, y<0$): $\text{Arg}(z) = -(\pi - \alpha) = -\pi + \alpha$
   • Quadrant IV ($x>0, y<0$): $\text{Arg}(z) = -\alpha$

2. Why is De Moivre's Theorem multi-valued when dealing with fractional exponents?

When $n = \frac{p}{q}$ is a rational fraction, $(\cos\theta + i\sin\theta)^{p/q}$ yields exactly $q$ distinct values. This non-uniqueness arises because the angle $\theta$ in polar form is periodic with period $2k\pi$. Expanding to $\frac{\theta + 2k\pi}{q}$ produces distinct points on the unit circle for $k = 0, 1, 2, \dots, q-1$. Beyond $k = q-1$, the geometric points repeat.

3. What is the fundamental difference between expanding $\cos(n\theta)$ and $\cos^n\theta$?

• Expanding $\cos(n\theta)$ in terms of powers of $\cos\theta$: Express $(\cos\theta + i\sin\theta)^n$ using the Binomial Theorem, apply De Moivre's Theorem to equate real parts, and substitute $\sin^2\theta = 1 - \cos^2\theta$.
• Expanding $\cos^n\theta$ in terms of multiple angles $\cos(k\theta)$: Set $2\cos\theta = x + x^{-1}$, expand $(x + x^{-1})^n$ using the Binomial Theorem, group symmetric terms $(x^k + x^{-k})$, and substitute $x^k + x^{-k} = 2\cos(k\theta)$.