S.M. Yusuf Mathematical Methods Chapter 1 Complex Numbers & De Moivre Theorem Solved Exercises PDF Guide | BS/BSc Math PU, UOS, QAU, KU & FBISE
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 $\text{Im}(z) = y$.
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:
$$z = r e^{i\theta}$$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 $\cos\left(\frac{p\theta}{q}\right) + i\sin\left(\frac{p\theta}{q}\right)$.
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 $\frac{2\pi}{n}$.
| 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 |
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.
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.
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 $32i$.
$$\sin^5\theta = \frac{1}{16} \left[ \sin(5\theta) - 5\sin(3\theta) + 10\sin\theta \right]$$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.
Q1: What is the exact value of $(1 + i)^8$?
Q2: The sum of all $n$-th roots of unity is always equal to:
Q3: The Principal Argument $\text{Arg}(z)$ of the complex number $z = -1 - i\sqrt{3}$ is:
Q4: By applying De Moivre's Theorem, $(\cos\theta - i\sin\theta)^{-4}$ simplifies to:
Q5: The geometric locus represented by $|z - 3 + 4i| = 5$ in the Argand plane is a:
Q6: The product of all $n$-th roots of unity is equal to:
Q7: If $x + \frac{1}{x} = 2\cos\theta$, then $x^n + \frac{1}{x^n}$ is equal to:
Q8: How many distinct complex roots exist for the equation $z^4 + 16 = 0$?
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:
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.