BS Math Mathematical Methods Chapter 1 Complex Numbers & De Moivre Theorem Solved Exercises PDF S.M. Yusuf | Punjab University, QAU, UOS, FBISE Guide
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.
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 $-\pi < \theta \le \pi$.
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:
$$\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$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 ($n \in \mathbb{N}$):
By the Principle of Mathematical Induction, De Moivre's Theorem holds for all positive integers $n$.
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:
$$z_k = R^{1/n} \left[ \cos\left(\frac{\phi + 2k\pi}{n}\right) + i\sin\left(\frac{\phi + 2k\pi}{n}\right) \right], \quad k = 0, 1, 2, \dots, n-1$$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 $\sin\theta$.
| 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. |
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.
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 $k = 0, 1, 2, 3, 4, 5$.
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:
$$\sin^6\theta = \frac{1}{32} \left( 10 - 15\cos(2\theta) + 6\cos(4\theta) - \cos(6\theta) \right)$$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 $k$:
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.
Q1: What is the principal value of $\arg(z)$ for $z = -1 - i$?
Q2: If $\omega$ is a complex cube root of unity, then what is the value of $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5$?
Q3: According to De Moivre's Theorem, $(\cos\theta - i\sin\theta)^{-n}$ is equal to:
Q4: Sum of the $n$-th roots of unity is always equal to:
Q5: The value of $\left(\frac{1 + i}{\sqrt{2}}\right)^8 + \left(\frac{1 - i}{\sqrt{2}}\right)^8$ is:
Q6: If $x = \cos\theta + i\sin\theta$, then $x^n - \frac{1}{x^n}$ equals:
Q7: All $n$-th roots of unity lie on a unit circle in the Argand plane and form vertices of a:
Q8: If $\arg(z) > 0$, then $\arg(-z) - \arg(z)$ is always equal to:
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.
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 $q$.
Use the following operational heuristic based on problem goals: