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

Theoretical Foundations & Core Theorems

Complex numbers extend the real number system $\mathbb{R}$ to a two-dimensional algebraically closed field $\mathbb{C}$. Defined as ordered pairs of real numbers $(x, y)$ or expressions of the form $z = x + iy$ where $i = \sqrt{-1}$, complex numbers serve as the fundamental backbone of modern mathematical analysis, quantum mechanics, signal processing, and fluid dynamics.

1. Polar Form & Euler's Identity

By representing a complex number $z = x + iy$ on the Argand plane, we map it to polar coordinates $(r, \theta)$: $$x = r \cos\theta, \quad y = r \sin\theta$$ where $r = |z| = \sqrt{x^2 + y^2}$ is the modulus (magnitude), and $\theta = \arg(z) = \tan^{-1}\left(\frac{y}{x}\right)$ is the argument. To ensure uniqueness, the Principal Argument $\text{Arg}(z)$ is restricted to the interval $(-\pi, \pi]$.

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

2. De Moivre's Theorem

De Moivre's Theorem is a fundamental bridge connecting complex analysis with trigonometry. It is stated in two primary cases:

Case 1: Integer Index ($n \in \mathbb{Z}$)
For any real number $\theta$ and any integer $n$: $$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$

Case 2: Rational Index ($n \in \mathbb{Q}$)
If $n = \frac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q > 0$, then 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 general $q$ distinct values are obtained via: $$\left[\cos(\theta + 2k\pi) + i\sin(\theta + 2k\pi)\right]^{p/q} = \cos\left(\frac{p(\theta + 2k\pi)}{q}\right) + i\sin\left(\frac{p(\theta + 2k\pi)}{q}\right), \quad k = 0, 1, 2, \dots, q-1$$

3. Trigonometric Expansions & Multi-Angle Identities

De Moivre's Theorem permits the expansion of $\cos(n\theta)$ and $\sin(n\theta)$ in powers of $\cos\theta$ and $\sin\theta$ using the Binomial Theorem on $(\cos\theta + i\sin\theta)^n$, and conversely allows powers such as $\cos^n\theta$ and $\sin^n\theta$ to be rewritten as linear combinations of sines and cosines of multiple angles using $x = e^{i\theta} \implies x + \frac{1}{x} = 2\cos\theta$ and $x - \frac{1}{x} = 2i\sin\theta$.

4. Summation of Trigonometric Series ($C + iS$ Method)

Complex variables simplify the summation of real trigonometric series of the form $C = \sum a_n \cos(n\theta)$ and $S = \sum a_n \sin(n\theta)$. By forming the complex sum $C + iS = \sum a_n e^{in\theta}$, the series transforms into a standard geometric, exponential, or logarithmic complex power series, which can be evaluated closed-form and subsequently separated into real and imaginary parts.

Formula Summary & Quick Reference

Key Concept	Formulas / Operational Conditions	Primary Applications
Modulus & Argument	$|z| = \sqrt{x^2 + y^2}$, $\text{Arg}(z) = \theta \in (-\pi, \pi]$	Converting Cartesian to Polar form
De Moivre's Theorem	$(\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$	Simplifying complex powers, multi-angle expansions
$n$-th Roots of Complex Number	$z_k = r^{1/n} \left[ \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right]$, $k = 0, \dots, n-1$	Solving polynomial equations $z^n = w$
Power Identites	$x^k + \frac{1}{x^k} = 2\cos(k\theta)$, $x^k - \frac{1}{x^k} = 2i\sin(k\theta)$	Expanding $\cos^n\theta$ and $\sin^n\theta$
$C + iS$ Summation	$C + iS = \sum_{n=0}^{\infty} a_n e^{in\theta}$; Real part $= C$, Imaginary part $= S$	Summing trigonometric series in analysis

Step-by-Step Solved Board Exam Questions

Problem 1: Trigonometric Expansion using De Moivre's Theorem

Question (PU / UOS Past Papers): Prove that: $$\cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1$$

Solution:

By De Moivre's Theorem, we have: $$\cos 6\theta + i\sin 6\theta = (\cos\theta + i\sin\theta)^6$$

Expanding the right-hand side using the Binomial Theorem $(a+b)^6 = \sum_{k=0}^6 \binom{6}{k} a^{6-k} b^k$ where $a = \cos\theta$ and $b = i\sin\theta$:

$$(\cos\theta + i\sin\theta)^6 = \binom{6}{0}\cos^6\theta + \binom{6}{1}\cos^5\theta(i\sin\theta) + \binom{6}{2}\cos^4\theta(i\sin\theta)^2 + \binom{6}{3}\cos^3\theta(i\sin\theta)^3 + \binom{6}{4}\cos^2\theta(i\sin\theta)^4 + \binom{6}{5}\cos\theta(i\sin\theta)^5 + \binom{6}{6}(i\sin\theta)^6$$

Evaluating the powers of $i$ ($i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, i^6 = -1$) and binomial coefficients ($\binom{6}{0}=1, \binom{6}{1}=6, \binom{6}{2}=15, \binom{6}{3}=20, \binom{6}{4}=15, \binom{6}{5}=6, \binom{6}{6}=1$):

$$= \cos^6\theta + 6i\cos^5\theta\sin\theta - 15\cos^4\theta\sin^2\theta - 20i\cos^3\theta\sin^3\theta + 15\cos^2\theta\sin^4\theta + 6i\cos\theta\sin^5\theta - \sin^6\theta$$

Equating the real parts on both sides:

$$\cos 6\theta = \cos^6\theta - 15\cos^4\theta\sin^2\theta + 15\cos^2\theta\sin^4\theta - \sin^6\theta$$

Substitute $\sin^2\theta = 1 - \cos^2\theta$ throughout:

$$\cos 6\theta = \cos^6\theta - 15\cos^4\theta(1 - \cos^2\theta) + 15\cos^2\theta(1 - \cos^2\theta)^2 - (1 - \cos^2\theta)^3$$

Expand the terms:

• $-15\cos^4\theta(1 - \cos^2\theta) = -15\cos^4\theta + 15\cos^6\theta$
• $15\cos^2\theta(1 - 2\cos^2\theta + \cos^4\theta) = 15\cos^2\theta - 30\cos^4\theta + 15\cos^6\theta$
• $-(1 - 3\cos^2\theta + 3\cos^4\theta - \cos^6\theta) = -1 + 3\cos^2\theta - 3\cos^4\theta + \cos^6\theta$

Combining all real terms:

$$\cos 6\theta = (1 + 15 + 15 + 1)\cos^6\theta + (-15 - 30 - 3)\cos^4\theta + (15 + 3)\cos^2\theta - 1$$ $$\cos 6\theta = 32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1$$

Hence proved.

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Problem 2: Cyclic Sum Identity (S.M. Yusuf Ex 1.3)

Question: 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:

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

Consider the sum $a + b + c$:

$$a + b + c = (\cos\alpha + \cos\beta + \cos\gamma) + i(\sin\alpha + \sin\beta + \sin\gamma) = 0 + i(0) = 0$$

From classical algebra, if $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$.

Substituting $a, b, c$ into the algebraic identity:

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

The right-hand side becomes:

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

Equating LHS and RHS:

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

Equating real and imaginary parts separately yields:

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

Hence proved.

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Problem 3: Solving Roots of Complex Equations

Question (QAU / FBISE): Solve the equation $x^7 + 1 = 0$ using De Moivre's Theorem and show that the sum of all roots is zero.

Solution:

The given equation is $x^7 = -1$.

First, write $-1$ in polar form:

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

To find all 7 roots, express the polar angle in general form by adding $2k\pi$ ($k \in \mathbb{Z}$):

$$-1 = \cos(\pi + 2k\pi) + i\sin(\pi + 2k\pi) = \cos((2k+1)\pi) + i\sin((2k+1)\pi)$$

Applying De Moivre's Theorem for roots ($x = (-1)^{1/7}$):

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

Using Euler's notation, $x_k = e^{i\frac{(2k+1)\pi}{7}} = e^{i\frac{\pi}{7}} \cdot e^{i\frac{2k\pi}{7}}$.

Let $\omega = e^{i\frac{2\pi}{7}}$ and $\alpha = e^{i\frac{\pi}{7}}$. Then the roots form a geometric progression:

$$x_k = \alpha \omega^k \quad \text{for } k = 0, 1, \dots, 6$$

Sum of roots $S$:

$$S = \sum_{k=0}^6 x_k = \alpha \sum_{k=0}^6 \omega^k = \alpha \left( \frac{1 - \omega^7}{1 - \omega} \right)$$

Since $\omega = e^{i\frac{2\pi}{7}}$, we have $\omega^7 = (e^{i\frac{2\pi}{7}})^7 = e^{i 2\pi} = 1$.

Therefore, since $\omega \neq 1$:

$$S = \alpha \left( \frac{1 - 1}{1 - \omega} \right) = 0$$

Hence, all roots are derived, and their sum is zero.

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Problem 4: Summation of Infinite Trigonometric Series ($C + iS$)

Question: Sum the infinite series: $$C = 1 + r\cos\theta + r^2\cos 2\theta + r^3\cos 3\theta + \dots \quad (|r| < 1)$$

Solution:

Let the given series be $C$. Consider the corresponding sine series $S$:

$$S = r\sin\theta + r^2\sin 2\theta + r^3\sin 3\theta + \dots$$

Form the complex sum $C + iS$:

$$C + iS = 1 + r(\cos\theta + i\sin\theta) + r^2(\cos 2\theta + i\sin 2\theta) + r^3(\cos 3\theta + i\sin 3\theta) + \dots$$

By Euler's formula $e^{in\theta} = \cos n\theta + i\sin n\theta$:

$$C + iS = 1 + r e^{i\theta} + r^2 e^{i2\theta} + r^3 e^{i3\theta} + \dots = 1 + (r e^{i\theta}) + (r e^{i\theta})^2 + (r e^{i\theta})^3 + \dots$$

This is an infinite geometric series with first term $a = 1$ and common ratio $z = r e^{i\theta}$. Since $|z| = |r e^{i\theta}| = |r| < 1$, the sum converges to $\frac{a}{1 - z}$:

$$C + iS = \frac{1}{1 - r e^{i\theta}} = \frac{1}{1 - r(\cos\theta + i\sin\theta)} = \frac{1}{(1 - r\cos\theta) - i r\sin\theta}$$

Rationalize the denominator by multiplying the numerator and denominator by the complex conjugate $(1 - r\cos\theta) + i r\sin\theta$:

$$C + iS = \frac{(1 - r\cos\theta) + i r\sin\theta}{[(1 - r\cos\theta) - i r\sin\theta][(1 - r\cos\theta) + i r\sin\theta]}$$ $$C + iS = \frac{(1 - r\cos\theta) + i r\sin\theta}{(1 - r\cos\theta)^2 + r^2\sin^2\theta}$$

Simplify the denominator:

$$(1 - r\cos\theta)^2 + r^2\sin^2\theta = 1 - 2r\cos\theta + r^2\cos^2\theta + r^2\sin^2\theta = 1 - 2r\cos\theta + r^2$$

Equating the real parts yields the value of $C$:

$$C = \frac{1 - r\cos\theta}{1 - 2r\cos\theta + r^2}$$

This completes the summation.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

Q1: Why must we convert $(\sin\theta + i\cos\theta)^n$ into $(\cos\phi + i\sin\phi)^n$ before applying De Moivre's Theorem?

De Moivre's Theorem is derived specifically from Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$. In the expression $\sin\theta + i\cos\theta$, the real part is a sine function and the imaginary part is a cosine function. Applying the exponent directly to angles without transformation leads to incorrect results. One must use co-function identities $\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)$ and $\cos\theta = \sin\left(\frac{\pi}{2} - \theta\right)$ to establish valid polar form standard identity before distributing $n$.

Q2: What is the primary difference between argument $\arg(z)$ and principal argument $\text{Arg}(z)$?

The argument $\arg(z)$ is a multi-valued function defined as $\arg(z) = \text{Arg}(z) + 2k\pi$ for $k \in \mathbb{Z}$, reflecting the periodic nature of trigonometric functions. Conversely, the principal argument $\text{Arg}(z)$ is uniquely restricted to a single half-open interval $(-\pi, \pi]$ (or $[0, 2\pi)$ in select university conventions). In Pakistani university examinations (PU, UOS, QAU), defaulting to $(-\pi, \pi]$ is required to avoid multi-valued ambiguities during logarithmic and fractional root calculations.

Q3: How do I handle negative signs inside expressions like $(\cos\theta - i\sin\theta)^n$?

Use the even/odd properties of trigonometric functions: $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$. Rewrite the expression as: $$\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)$$ Then apply De Moivre's Theorem: $$[\cos(-\theta) + i\sin(-\theta)]^n = \cos(-n\theta) + i\sin(-n\theta) = \cos(n\theta) - i\sin(n\theta)$$ Alternatively, treat it as $(\cos\theta + i\sin\theta)^{-n}$.