S.M. Yusuf Mathematical Methods Chapter 1 Complex Numbers & De Moivre Theorem Solved Exercises PDF Guide | BS Math PU UOS KU QAU & Board Exams
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:
$$z = x + i y \quad \text{where } x, y \in \mathbb{R} \text{ and } i = \sqrt{-1}$$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:
Euler's identity forms the connective tissue between complex exponentials and trigonometric functions. Formally defined via Maclaurin series expansions:
$$e^{i\theta} = \sum_{n=0}^{\infty} \frac{(i\theta)^n}{n!} = \left( 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \dots \right) + i \left( \theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \dots \right) = \cos\theta + i \sin\theta$$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):
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.
| 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 |
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. $\blacksquare$
---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. $\blacksquare$
---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 $k = 0, 1, 2, 3, 4, 5, 6$:
Final Solution Set: $S.S. = \left\{ \exp\left(i\frac{(2k+1)\pi}{7}\right) \;\middle|\; k = 0, 1, 2, 3, 4, 5, 6 \right\}$. $\blacksquare$
---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. $\blacksquare$
Q1: What is the principal argument $\text{Arg}(z)$ of the complex number $z = -1 - i\sqrt{3}$?
Q2: According to De Moivre's Theorem, what is the value of $(\cos\theta - i\sin\theta)^{-5}$?
Q3: What is the sum of all $n$-th roots of unity for any integer $n > 1$?
Q4: What is the product of all $n$-th roots of unity?
Q5: If $x = e^{i\theta}$, what does $x^n - \frac{1}{x^n}$ evaluate to?
Q6: The locus of points satisfying $|z - 3| = |z + 3|$ in the complex plane represents a:
Q7: What is the simplified modulus $|e^{z}|$ when $z = x + iy$?
Q8: If $\omega$ is a complex cube root of unity, what is the value of $(1 - \omega + \omega^2)^5 + (1 + \omega - \omega^2)^5$?
To avoid common sign errors when computing $\theta = \text{Arg}(z)$ for $z = x + iy$:
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.