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

1. Theoretical Foundations & Core Theorems

Complex numbers form the mathematical bedrock of advanced analysis, linear algebra, signal processing, and quantum mechanics. A complex number $z$ is formally defined as an ordered pair of real numbers $(x, y) \in \mathbb{R}^2$, expressed in algebraic (Cartesian) form as:

$$z = x + iy$$

where $i = \sqrt{-1}$ represents the imaginary unit. The real part is denoted by $\text{Re}(z) = x$ and the imaginary part by $\text{Im}(z) = y$.

Polar and Exponential Representations

By transforming Cartesian coordinates $(x, y)$ into polar coordinates $(r, \theta)$, where $r = |z| = \sqrt{x^2 + y^2}$ is the modulus and $\theta = \text{arg}(z) = \arctan\left(\frac{y}{x}\right)$ is the argument (angle with the positive real axis), we express $z$ as:

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

Using Euler's Formula, $e^{i\theta} = \cos\theta + i\sin\theta$, any complex number can be compactly written in exponential form:

$$z = r e^{i\theta}$$

De Moivre's Theorem

Theorem Statement: For any real number $n$ (integer or rational fraction),

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

Case 1 (Integral Exponent): If $n \in \mathbb{Z}^+$, the theorem is readily proved via Mathematical Induction. If $n \in \mathbb{Z}^-$, let $n = -m$ where $m \in \mathbb{Z}^+$. Then:

$$(\cos\theta + i\sin\theta)^{-m} = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos(m\theta) + i\sin(m\theta)} = \cos(m\theta) - i\sin(m\theta) = \cos(-m\theta) + i\sin(-m\theta)$$

Case 2 (Rational Exponent): If $n = \frac{p}{q}$ where $p, q \in \mathbb{Z}, q > 0$, then one of the $q$ values of $(\cos\theta + i\sin\theta)^{p/q}$ is $\cos\left(\frac{p\theta}{q}\right) + i\sin\left(\frac{p\theta}{q}\right)$.

Finding Roots of Complex Numbers

To determine all $n$-th roots of a complex number $z = r(\cos\theta + i\sin\theta)$, we leverage the periodicity of trigonometric functions ($2k\pi$ added to the argument):

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

This generates exactly $n$ distinct roots distributed uniformly around a circle of radius $r^{1/n}$ in the Argand plane, separated by equal angular increments of $\frac{2\pi}{n}$.

Real-World & University Application Context

• AC Circuit Analysis: Impedance $Z = R + iX$ utilizes complex polar algebra to combine magnitude and phase shift.
• Quantum Wave Mechanics: Schrödinger's wave equation $\hat{H}\Psi = i\hbar \frac{\partial \Psi}{\partial t}$ inherently relies on complex phase vectors.
• Fourier Analysis: Decomposition of periodic signals into complex exponential harmonics $e^{i n \omega t}$.

2. Formula Summary & Quick Reference

Concept / Topic	Mathematical Formula / Condition	Academic Application / Context
Modulus & Argument	$|z| = \sqrt{x^2+y^2}, \quad \text{Arg}(z) = \text{Arg}(x+iy) \in (-\pi, \pi]$	Converting Cartesian form to Polar/Euler form
Euler's Identity	$e^{i\pi} + 1 = 0 \implies e^{i\theta} = \cos\theta + i\sin\theta$	Foundational identity in complex analysis
De Moivre's Theorem	$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$	Simplifying powers and expansions of complex terms
Roots of Unity	$w_k = e^{i\frac{2k\pi}{n}}, \quad k=0,1,\dots,n-1$	Solving equations of the form $z^n = 1$
Sum of $n$-th Roots	$\sum_{k=0}^{n-1} e^{i\frac{2k\pi}{n}} = 0$	Geometric symmetry of roots on Argand diagram
Product of $n$-th Roots	$\prod_{k=0}^{n-1} w_k = (-1)^{n-1}$	Algebraic root properties in polynomial theory
Expansion of $\cos^n\theta, \sin^n\theta$	$x = e^{i\theta} \implies x + \frac{1}{x} = 2\cos\theta, \quad x - \frac{1}{x} = 2i\sin\theta$	Transforming powers into multiple angle sines/cosines
Circle Locus	$|z - z_0| = R$	Geometric locus representation in complex plane

3. Step-by-Step Solved Board Exam Questions

Problem 1 (S.M. Yusuf Exercise 1.2 Classic Exam Problem)

Question: Prove that:

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

Detailed Solution:

Step 1: Express terms using co-function identities.

Recall that $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ and $\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)$. Let $\phi = \frac{\pi}{2} - \theta$.

Substituting these into the expression:

$$\text{Numerator} = 1 + \cos\phi + i\sin\phi$$ $$\text{Denominator} = 1 + \cos\phi - i\sin\phi$$

Step 2: Apply half-angle trigonometric formulas.

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

$$\text{Numerator} = 2\cos^2\left(\frac{\phi}{2}\right) + 2i\sin\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right) = 2\cos\left(\frac{\phi}{2}\right) \left[ \cos\left(\frac{\phi}{2}\right) + i\sin\left(\frac{\phi}{2}\right) \right]$$ $$\text{Denominator} = 2\cos^2\left(\frac{\phi}{2}\right) - 2i\sin\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right) = 2\cos\left(\frac{\phi}{2}\right) \left[ \cos\left(\frac{\phi}{2}\right) - i\sin\left(\frac{\phi}{2}\right) \right]$$

Step 3: Simplify the ratio inside the power.

$$\frac{\text{Numerator}}{\text{Denominator}} = \frac{\cos\left(\frac{\phi}{2}\right) + i\sin\left(\frac{\phi}{2}\right)}{\cos\left(\frac{\phi}{2}\right) - i\sin\left(\frac{\phi}{2}\right)} = \frac{e^{i\phi/2}}{e^{-i\phi/2}} = e^{i\phi} = \cos\phi + i\sin\phi$$

Step 4: Raise to the power $n$ using De Moivre's Theorem.

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

Substitute back $\phi = \frac{\pi}{2} - \theta$:

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

Hence Proved.

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Problem 2 (Solving Complex Equations using De Moivre's Theorem)

Question: Solve the equation $z^7 + 1 = 0$ using De Moivre's Theorem and state all roots in polar form.

Detailed Solution:

Step 1: Isolate $z^7$ and write the RHS in polar form.

$$z^7 = -1$$

The modulus of $-1$ is $r = 1$, and its principal argument is $\theta = \pi$.

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

Step 2: Introduce general polar angle by adding $2k\pi$.

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

Step 3: Apply De Moivre's Theorem for fractional exponents.

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

Step 4: Explicitly list the seven distinct roots.

• 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 $k=3$: $z_3 = \cos\left(\pi\right) + i\sin\left(\pi\right) = -1$
• 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 $k=6$: $z_6 = \cos\left(\frac{13\pi}{7}\right) + i\sin\left(\frac{13\pi}{7}\right) = e^{i 13\pi/7}$

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Problem 3 (Trigonometric Power Expansion)

Question: Expand $\sin^5\theta$ in a series of sines of multiples of $\theta$.

Detailed Solution:

Step 1: Define complex variable relation.

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

Subtracting gives: $x - \frac{1}{x} = 2i\sin\theta$.

Also recall that $x^k - \frac{1}{x^k} = 2i\sin(k\theta)$ for any integer $k$.

Step 2: Raise both sides to the 5th power.

$$(2i\sin\theta)^5 = \left(x - \frac{1}{x}\right)^5$$ $$32 i \sin^5\theta = \left(x - \frac{1}{x}\right)^5$$

Step 3: Apply Binomial Expansion.

$$\left(x - \frac{1}{x}\right)^5 = x^5 - 5x^4\left(\frac{1}{x}\right) + 10x^3\left(\frac{1}{x^2}\right) - 10x^2\left(\frac{1}{x^3}\right) + 5x\left(\frac{1}{x^4}\right) - \frac{1}{x^5}$$ $$= x^5 - 5x^3 + 10x - \frac{10}{x} + \frac{5}{x^3} - \frac{1}{x^5}$$

Step 4: Group corresponding symmetric powers.

$$32i \sin^5\theta = \left(x^5 - \frac{1}{x^5}\right) - 5\left(x^3 - \frac{1}{x^3}\right) + 10\left(x - \frac{1}{x}\right)$$

Step 5: Substitute $x^k - \frac{1}{x^k} = 2i\sin(k\theta)$.

$$32i \sin^5\theta = 2i\sin(5\theta) - 5(2i\sin(3\theta)) + 10(2i\sin\theta)$$ $$32i \sin^5\theta = 2i \left[ \sin(5\theta) - 5\sin(3\theta) + 10\sin\theta \right]$$

Step 6: Divide both sides by $32i$.

$$\sin^5\theta = \frac{1}{16} \left[ \sin(5\theta) - 5\sin(3\theta) + 10\sin\theta \right]$$

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Problem 4 (Separation of Hyperbolic & Trigonometric Complex Functions)

Question: If $\sin(\alpha + i\beta) = x + iy$, prove that:

$$\text{(i) } \frac{x^2}{\cosh^2\beta} + \frac{y^2}{\sinh^2\beta} = 1 \qquad \text{and} \qquad \text{(ii) } \frac{x^2}{\sin^2\alpha} - \frac{y^2}{\cos^2\alpha} = 1$$

Detailed Solution:

Step 1: Expand $\sin(\alpha + i\beta)$ using sine addition identity.

$$\sin(\alpha + i\beta) = \sin\alpha\cos(i\beta) + \cos\alpha\sin(i\beta)$$

Using identities $\cos(i\beta) = \cosh\beta$ and $\sin(i\beta) = i\sinh\beta$:

$$x + iy = \sin\alpha\cosh\beta + i\cos\alpha\sinh\beta$$

Step 2: Equate real and imaginary parts.

$$x = \sin\alpha\cosh\beta \implies \sin\alpha = \frac{x}{\cosh\beta}, \quad \cosh\beta = \frac{x}{\sin\alpha}$$ $$y = \cos\alpha\sinh\beta \implies \cos\alpha = \frac{y}{\sinh\beta}, \quad \sinh\beta = \frac{y}{\cos\alpha}$$

Step 3: Prove Part (i).

Using the fundamental trigonometric identity $\sin^2\alpha + \cos^2\alpha = 1$:

$$\left(\frac{x}{\cosh\beta}\right)^2 + \left(\frac{y}{\sinh\beta}\right)^2 = 1 \implies \frac{x^2}{\cosh^2\beta} + \frac{y^2}{\sinh^2\beta} = 1$$

Step 4: Prove Part (ii).

Using the fundamental hyperbolic identity $\cosh^2\beta - \sinh^2\beta = 1$:

$$\left(\frac{x}{\sin\alpha}\right)^2 - \left(\frac{y}{\cos\alpha}\right)^2 = 1 \implies \frac{x^2}{\sin^2\alpha} - \frac{y^2}{\cos^2\alpha} = 1$$

Hence Proved.

4. Interactive Practice Quiz (Clickable MCQs)

5. Frequently Asked Questions (FAQs)

Q1: Why must a complex number be converted to exact standard polar form $\cos\theta + i\sin\theta$ before applying De Moivre's Theorem?

De Moivre's Theorem is derived from Euler's identity $e^{i\theta} = \cos\theta + i\sin\theta$. If a complex expression contains a minus sign between terms (e.g., $\cos\theta - i\sin\theta$) or swapped components (e.g., $\sin\theta + i\cos\theta$), it does not directly match $e^{i\theta}$. You must first transform it into standard polar form using co-function identities or angle sign changes:

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

Applying the theorem without this conversion yields incorrect sign and phase values.

Q2: What is the primary distinction between the General Argument $\text{arg}(z)$ and the Principal Argument $\text{Arg}(z)$?

The argument of a complex number is multi-valued due to the $2\pi$ periodicity of trigonometric functions:

$$\text{arg}(z) = \text{Arg}(z) + 2k\pi, \quad k \in \mathbb{Z}$$

The Principal Argument $\text{Arg}(z)$ is restricted to a single continuous interval of length $2\pi$, defined rigorously by international standard as:

$$-\pi < \text{Arg}(z) \le \pi$$

In Pakistani university examinations (PU, UOS, QAU), failing to restrict the final angle of a root or modulus-argument form to $(-\pi, \pi]$ results in marks deduction.

Q3: How do we avoid common errors when computing roots of negative real or purely imaginary numbers?

Students frequently forget to convert negative real numbers or pure imaginary numbers into full polar form prior to taking roots. Always write:

• For real negative numbers (e.g., $-a$ where $a > 0$): $-a = a(\cos\pi + i\sin\pi) = a e^{i(\pi + 2k\pi)}$
• For positive imaginary numbers (e.g., $+ia$): $ia = a\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = a e^{i(\frac{\pi}{2} + 2k\pi)}$
• For negative imaginary numbers (e.g., $-ia$): $-ia = a\left(\cos\left(-\frac{\pi}{2}\right) + i\sin\left(-\frac{\pi}{2}\right)\right) = a e^{i(-\frac{\pi}{2} + 2k\pi)}$

Adding the $2k\pi$ periodicity term before dividing by the root power $n$ guarantees that all $n$ unique non-repeating roots (for $k = 0, 1, \dots, n-1$) are correctly obtained.