BS Math Mathematical Methods Chapter 1 Complex Numbers & De Moivre Theorem Solved Exercises PDF S.M. Yusuf | Punjab University, QAU, UOS, FBISE Guide

Theoretical Foundations & Core Theorems

The study of complex numbers forms the bedrock of advanced analysis, linear algebra, and mathematical physics. In S.M. Yusuf's Mathematical Methods (Chapter 1), complex numbers are developed beyond simple algebraic structures into powerful analytical tools using geometric and trigonometric representations.

1. Algebraic and Polar Representation of Complex Numbers

A complex number $z$ is expressed in Cartesian form as $z = x + iy$, where $x = \text{Re}(z) \in \mathbb{R}$, $y = \text{Im}(z) \in \mathbb{R}$, and $i = \sqrt{-1}$. By introducing polar coordinates $(r, \theta)$ via the transformations $x = r \cos\theta$ and $y = r \sin\theta$, we obtain the polar form:

$$z = r (\cos\theta + i\sin\theta)$$

where $r = |z| = \sqrt{x^2 + y^2}$ represents the non-negative magnitude (modulus) of $z$, and $\theta = \arg(z) = \tan^{-1}\left(\frac{y}{x}\right)$ is the argument of $z$. The principal argument, denoted by $\text{Arg}(z)$, is restricted to the interval $-\pi < \theta \le \pi$.

2. Euler's Formula and Exponential Form

Using the Taylor series expansions of $e^x$, $\cos x$, and $\sin x$, Euler established the fundamental identity:

$$e^{i\theta} = \cos\theta + i\sin\theta$$

Consequently, any complex number can be compactly written in exponential form as $z = r e^{i\theta}$. From Euler's formula, the trigonometric functions can be isolated as:

$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$

3. De Moivre's Theorem

Theorem Statement: For any integer $n \in \mathbb{Z}$,

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

For any rational number $n = \frac{p}{q}$ (where $q > 0$ and $\gcd(p,q) = 1$), one of the values of $(\cos\theta + i\sin\theta)^n$ is $\cos(n\theta) + i\sin(n\theta)$.

Proof Strategy for Positive Integers ($n \in \mathbb{N}$):

• Base Step ($n=1$): $(\cos\theta + i\sin\theta)^1 = \cos(1\cdot\theta) + i\sin(1\cdot\theta)$, which holds trivially.
• Inductive Hypothesis: Assume true for $n = k$: $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$.
• Inductive Step ($n = k+1$): $$\begin{aligned} (\cos\theta + i\sin\theta)^{k+1} &= (\cos\theta + i\sin\theta)^k (\cos\theta + i\sin\theta) \\ &= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta) \\ &= (\cos k\theta \cos\theta - \sin k\theta \sin\theta) + i(\sin k\theta \cos\theta + \cos k\theta \sin\theta) \\ &= \cos((k+1)\theta) + i\sin((k+1)\theta) \end{aligned}$$

By the Principle of Mathematical Induction, De Moivre's Theorem holds for all positive integers $n$.

4. Roots of Complex Numbers

To solve equations of the form $z^n = w$, where $w = R(\cos\phi + i\sin\phi)$, we express $w$ in its periodic polar form: $w = R \left(\cos(\phi + 2k\pi) + i\sin(\phi + 2k\pi)\right)$ for $k \in \mathbb{Z}$. Applying De Moivre's Theorem for fractional exponents gives the $n$ distinct roots:

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

5. Expansion Techniques ($\cos^n\theta, \sin^n\theta$ and $\cos(n\theta), \sin(n\theta)$)

Let $x = e^{i\theta} = \cos\theta + i\sin\theta$. Then $x^{-1} = e^{-i\theta} = \cos\theta - i\sin\theta$. This leads to two critical operational identities:

$$x + \frac{1}{x} = 2\cos\theta, \quad x - \frac{1}{x} = 2i\sin\theta$$ $$x^k + \frac{1}{x^k} = 2\cos(k\theta), \quad x^k - \frac{1}{x^k} = 2i\sin(k\theta)$$

Expanding $(x + x^{-1})^n$ or $(x - x^{-1})^n$ via the Binomial Theorem yields powers of trigonometric functions in terms of multiple angles. Conversely, expanding $(\cos\theta + i\sin\theta)^n$ using the Binomial Theorem and equating real and imaginary parts expresses multiple-angle trigonometric functions $\cos(n\theta)$ and $\sin(n\theta)$ as powers of $\cos\theta$ and $\sin\theta$.

Formula Summary & Quick Reference

Key Concept	Formula / Mathematical Identity	Primary Application / Context
Modulus & Argument	$|z| = \sqrt{x^2+y^2}$, $\theta = \tan^{-1}\left(\frac{y}{x}\right)$	Converting Cartesian coordinates to Polar form.
De Moivre's Identity	$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$	Simplifying high powers of complex numbers.
Euler's Form	$z = r e^{i\theta}$ where $e^{i\theta} = \cos\theta + i\sin\theta$	Complex multiplication, division, and logarithm operations.
$n$-th Roots of Unity	$\omega_k = e^{i\frac{2k\pi}{n}} = \cos\left(\frac{2k\pi}{n}\right) + i\sin\left(\frac{2k\pi}{n}\right)$	Solving $z^n = 1$; roots form a cyclic group of order $n$.
General $n$-th Roots	$z_k = r^{1/n} e^{i\left(\frac{\theta + 2k\pi}{n}\right)}, \quad k=0,1,\dots,n-1$	Solving polynomial equations of the form $z^n = w$.
Power Expansion Identities	$2\cos\theta = x + x^{-1}$, $2i\sin\theta = x - x^{-1}$	Expressing $\cos^n\theta$ or $\sin^n\theta$ as linear combinations of multiple angles.
Multiple-Angle Identities	$x^k + x^{-k} = 2\cos(k\theta)$, $x^k - x^{-k} = 2i\sin(k\theta)$	Integrating high powers of trigonometric functions.

Step-by-Step Solved Board Exam Questions

Problem 1: Expansion of Multiple Angles

Question: Use De Moivre's Theorem to express $\cos(5\theta)$ in terms of powers of $\cos\theta$, and show that:

$$\frac{\sin(5\theta)}{\sin\theta} = 16\cos^4\theta - 12\cos^2\theta + 1 \quad (\text{for } \sin\theta \neq 0)$$

Solution:

Step 1: Apply De Moivre's Theorem for $n = 5$:

$$\cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5$$

Step 2: Expand the right-hand side using the Binomial Theorem $(a+b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5$ with $a = \cos\theta$ and $b = i\sin\theta$:

$$\begin{aligned} (\cos\theta + i\sin\theta)^5 = \cos^5\theta &+ 5\cos^4\theta(i\sin\theta) + 10\cos^3\theta(i\sin\theta)^2 \\ &+ 10\cos^2\theta(i\sin\theta)^3 + 5\cos\theta(i\sin\theta)^4 + (i\sin\theta)^5 \end{aligned}$$

Substituting powers of $i$ ($i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i$):

$$\begin{aligned} (\cos\theta + i\sin\theta)^5 = \cos^5\theta &+ 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta \\ &- 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta \end{aligned}$$

Grouping into real and imaginary components:

$$\begin{aligned} (\cos\theta + i\sin\theta)^5 = &\left(\cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta\right) \\ &+ i\left(5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta\right) \end{aligned}$$

Step 3: Equate real parts to find $\cos(5\theta)$:

$$\cos(5\theta) = \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - \cos^2\theta)^2$$ $$\cos(5\theta) = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta)$$ $$\cos(5\theta) = 11\cos^5\theta - 10\cos^3\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta$$ $$\cos(5\theta) = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta$$

Step 4: Equate imaginary parts to find $\sin(5\theta)$:

$$\sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta$$

Divide both sides by $\sin\theta$ ($\sin\theta \neq 0$):

$$\frac{\sin(5\theta)}{\sin\theta} = 5\cos^4\theta - 10\cos^2\theta\sin^2\theta + \sin^4\theta$$

Substitute $\sin^2\theta = 1 - \cos^2\theta$ and $\sin^4\theta = (1 - \cos^2\theta)^2 = 1 - 2\cos^2\theta + \cos^4\theta$:

$$\begin{aligned} \frac{\sin(5\theta)}{\sin\theta} &= 5\cos^4\theta - 10\cos^2\theta(1 - \cos^2\theta) + (1 - 2\cos^2\theta + \cos^4\theta) \\ &= 5\cos^4\theta - 10\cos^2\theta + 10\cos^4\theta + 1 - 2\cos^2\theta + \cos^4\theta \\ &= (5 + 10 + 1)\cos^4\theta - (10 + 2)\cos^2\theta + 1 \\ &= 16\cos^4\theta - 12\cos^2\theta + 1 \end{aligned}$$

Q.E.D.

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Problem 2: Solving Polynomial Equations with Complex Roots

Question: Solve the equation $(x + 1)^6 + (x - 1)^6 = 0$ using De Moivre's Theorem.

Solution:

Step 1: Rearrange the given equation:

$$(x + 1)^6 = -(x - 1)^6$$

Dividing by $(x - 1)^6$ (assuming $x \neq 1$):

$$\left(\frac{x+1}{x-1}\right)^6 = -1$$

Step 2: Express $-1$ in general polar form:

$$-1 = \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) = e^{i(2k+1)\pi}, \quad k = 0, 1, 2, 3, 4, 5$$

Step 3: Take the 6th root on both sides:

$$\frac{x+1}{x-1} = \left[\cos(2k+1)\pi + i\sin(2k+1)\pi\right]^{1/6}$$

Applying De Moivre's Theorem:

$$\frac{x+1}{x-1} = \cos\left(\frac{(2k+1)\pi}{6}\right) + i\sin\left(\frac{(2k+1)\pi}{6}\right)$$

Let $\alpha = \frac{(2k+1)\pi}{6}$. Thus, $\frac{x+1}{x-1} = \cos\alpha + i\sin\alpha = e^{i\alpha}$.

Step 4: Apply Componendo and Dividendo to solve for $x$:

$$\frac{(x+1) + (x-1)}{(x+1) - (x-1)} = \frac{e^{i\alpha} + 1}{e^{i\alpha} - 1}$$ $$\frac{2x}{2} = \frac{e^{i\alpha} + 1}{e^{i\alpha} - 1}$$ $$x = \frac{e^{i\alpha/2}(e^{i\alpha/2} + e^{-i\alpha/2})}{e^{i\alpha/2}(e^{i\alpha/2} - e^{-i\alpha/2})} = \frac{2\cos(\alpha/2)}{2i\sin(\alpha/2)}$$ $$x = \frac{1}{i} \cot\left(\frac{\alpha}{2}\right) = -i \cot\left(\frac{\alpha}{2}\right)$$

Step 5: Substitute $\alpha = \frac{(2k+1)\pi}{6}$ back into the solution:

$$x_k = -i \cot\left(\frac{(2k+1)\pi}{12}\right), \quad \text{for } k = 0, 1, 2, 3, 4, 5$$

Thus, the 6 distinct imaginary roots are given for $k = 0, 1, 2, 3, 4, 5$.

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Problem 3: Expanding Powers of Sine in Terms of Multiple Angles

Question: Express $\sin^6\theta$ as a series of cosines of multiples of $\theta$.

Solution:

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

$$2i\sin\theta = x - \frac{1}{x}$$

Step 2: Raise both sides to the 6th power:

$$(2i\sin\theta)^6 = \left(x - \frac{1}{x}\right)^6$$ $$2^6 i^6 \sin^6\theta = -64 \sin^6\theta \quad (\text{since } i^6 = (i^2)^3 = (-1)^3 = -1)$$

Step 3: Expand $(x - x^{-1})^6$ using the Binomial Theorem:

$$\left(x - \frac{1}{x}\right)^6 = \binom{6}{0}x^6 - \binom{6}{1}x^4 + \binom{6}{2}x^2 - \binom{6}{3}x^0 + \binom{6}{4}x^{-2} - \binom{6}{5}x^{-4} + \binom{6}{6}x^{-6}$$ $$\left(x - \frac{1}{x}\right)^6 = x^6 - 6x^4 + 15x^2 - 20 + 15x^{-2} - 6x^{-4} + x^{-6}$$

Step 4: Group terms with positive and negative powers:

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

Step 5: Substitute the identity $x^k + x^{-k} = 2\cos(k\theta)$:

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

Step 6: Divide by $-64$ to isolate $\sin^6\theta$:

$$\sin^6\theta = -\frac{2}{64}\cos(6\theta) + \frac{12}{64}\cos(4\theta) - \frac{30}{64}\cos(2\theta) + \frac{20}{64}$$ $$\sin^6\theta = -\frac{1}{32}\cos(6\theta) + \frac{3}{16}\cos(4\theta) - \frac{15}{32}\cos(2\theta) + \frac{5}{16}$$

Rearranging in standard descending order:

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

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Problem 4: Roots of Complex Numbers and Geometric Evaluation

Question: Find all values of $(1 + i\sqrt{3})^{3/4}$ and show that the product of all these values is $8$.

Solution:

Step 1: Convert $z = 1 + i\sqrt{3}$ to polar form.

Modulus: $r = |z| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1+3} = 2$.

Argument: $\theta = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3}$.

Thus, $1 + i\sqrt{3} = 2 e^{i\left(\frac{\pi}{3} + 2k\pi\right)}$.

Step 2: Compute $w = z^3 = (1 + i\sqrt{3})^3$:

$$w = 2^3 e^{i 3\left(\frac{\pi}{3} + 2k\pi\right)} = 8 e^{i(\pi + 6k\pi)} = 8 e^{i\pi(2m+1)}$$

In general polar form: $w = 8 \left[ \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) \right]$.

Step 3: Take the 4th root $(w)^{1/4}$ using De Moivre's Theorem:

$$w_k = 8^{1/4} \left[ \cos\left(\frac{\pi + 2k\pi}{4}\right) + i\sin\left(\frac{\pi + 2k\pi}{4}\right) \right], \quad k = 0, 1, 2, 3$$ $$w_k = 2^{3/4} e^{i\frac{(2k+1)\pi}{4}}, \quad k = 0, 1, 2, 3$$

Evaluating for each $k$:

• For $k = 0$: $w_0 = 2^{3/4} e^{i\pi/4}$
• For $k = 1$: $w_1 = 2^{3/4} e^{i3\pi/4}$
• For $k = 2$: $w_2 = 2^{3/4} e^{i5\pi/4}$
• For $k = 3$: $w_3 = 2^{3/4} e^{i7\pi/4}$

Step 4: Calculate the product $P = w_0 \cdot w_1 \cdot w_2 \cdot w_3$:

$$P = \left(2^{3/4}\right)^4 \cdot e^{i\left(\frac{\pi}{4} + \frac{3\pi}{4} + \frac{5\pi}{4} + \frac{7\pi}{4}\right)}$$ $$P = 2^3 \cdot e^{i\left(\frac{16\pi}{4}\right)} = 8 \cdot e^{i 4\pi}$$

Since $e^{i 4\pi} = \cos(4\pi) + i\sin(4\pi) = 1 + i(0) = 1$:

$$P = 8 \times 1 = 8$$

Hence proved.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

1. What is the fundamental difference between the Principal Argument $\text{Arg}(z)$ and the General Argument $\arg(z)$?

The argument of a complex number is multi-valued due to the $2\pi$ periodicity of trigonometric functions. The general argument is defined as $\arg(z) = \text{Arg}(z) + 2k\pi$ for $k \in \mathbb{Z}$. In contrast, the Principal Argument $\text{Arg}(z)$ is uniquely restricted to the single interval $(-\pi, \pi]$. When evaluating fractional powers using De Moivre's Theorem, one must use the general argument $2k\pi + \theta$ to obtain all distinct roots before isolating specific solutions.

2. Why does caution need to be applied when applying De Moivre's Theorem to non-integer exponents?

For integer exponents $n \in \mathbb{Z}$, the expression $(\cos\theta + i\sin\theta)^n$ yields a single unique value. However, when $n$ is a rational number $\frac{p}{q}$ (in lowest terms), $(\cos\theta + i\sin\theta)^{p/q}$ represents a multi-valued function with exactly $q$ distinct complex values. Applying De Moivre's naive formula $\cos(n\theta) + i\sin(n\theta)$ yields only one of these values (the principal root). To capture all roots, the periodic term $2k\pi$ must be added to the argument prior to dividing by $q$.

3. How do I decide whether to expand via Binomial Theorem or use $x + x^{-1} = 2\cos\theta$?

Use the following operational heuristic based on problem goals:

• Expressing multiple angles ($\cos(n\theta), \sin(n\theta)$) in terms of powers ($\cos^n\theta$): Expand $(\cos\theta + i\sin\theta)^n$ using the Binomial Theorem, apply De Moivre's identity, and equate real/imaginary parts (e.g., Solved Problem 1).
• Expressing powers ($\cos^n\theta, \sin^n\theta$) in terms of linear multiple angles ($\cos(n\theta)$): Let $x = e^{i\theta}$, set up $(2\cos\theta)^n = (x + x^{-1})^n$ or $(2i\sin\theta)^n = (x - x^{-1})^n$, expand binomially, and combine symmetric terms using $x^k + x^{-k} = 2\cos(k\theta)$ (e.g., Solved Problem 3).