Further Complex Numbers

# Further Complex Numbers ## Objectifs et prérequis This chapter builds on foundational knowledge of complex numbers from Lower Sixth and aims to deepen understanding of their geometric and algebraic properties, particularly through polar and exponential forms, De Moivre’s theorem, and their applications in solving equations, analyzing loci, and interpreting plane transformations. Students should be proficient in: - Arithmetic with complex numbers in Cartesian form ($z = x + iy$) - Modulus and argument: $|z| = \sqrt{x^2 + y^2}$, $\arg(z) = \theta$ where $\tan\theta = y/x$ - Basic properties of conjugates: $\overline{z} = x - iy$, $z\overline{z} = |z|^2$ - Solving quadratic equations with complex roots - Basic matrix operations (addition, multiplication, determinant, inverse for 2×2) ## Introduction Further Complex Numbers extends the study of complex numbers beyond algebraic manipulation to their geometric interpretation in the Argand diagram and their utility in describing and analyzing plane transformations. By representing complex numbers in polar and exponential forms, we unlock powerful tools such as De Moivre’s theorem for computing powers and roots efficiently. This chapter further explores how complex equations define loci (lines, circles, arcs) in the complex plane, and how transformations like rotations, reflections, enlargements, and shears can be elegantly expressed and composed using complex arithmetic and 2×2 real matrices. The interplay between modulus inequalities, orientation-preserving/reversing motions, and similitudes provides a unified framework for understanding geometric transformations, essential for advanced mathematics and applications in physics and engineering. ## I. Forms of a Complex Number A complex number $z$ can be expressed in multiple equivalent forms, each useful in different contexts. ### Cartesian (Algebraic) Form $$ z = x + iy \quad \text{where } x, y \in \mathbb{R}, \; i^2 = -1 $$ Here, $x = \operatorname{Re}(z)$ is the real part, $y = \operatorname{Im}(z)$ is the imaginary part. > **To retain**: The Cartesian form is ideal for addition and subtraction: $(x_1 + iy_1) + (x_2 + iy_2) = (x_1 + x_2) + i(y_1 + y_2)$. ### Polar Form $$ z = r(\cos\theta + i\sin\theta) \quad \text{where } r = |z| \geq 0, \; \theta = \arg(z) $$ The modulus $r$ is the distance from the origin; the argument $\theta$ is the angle from the positive real axis, measured in radians. The principal argument $\operatorname{Arg}(z)$ satisfies $-\pi < \theta \leq \pi$. > **To retain**: Conversion formulas: $x = r\cos\theta$, $y = r\sin\theta$, $r = \sqrt{x^2 + y^2}$, $\theta = \tan^{-1}(y/x)$ (adjusted for quadrant). ### Exponential Form (Euler’s Formula) Using Euler’s identity $e^{i\theta} = \cos\theta + i\sin\theta$, we write: $$ z = re^{i\theta} $$ This form simplifies multiplication, division, and powers. > **To retain**: $e^{i\theta_1} \cdot e^{i\theta_2} = e^{i(\theta_1 + \theta_2)}$, so $z_1 z_2 = r_1 r_2 e^{i(\theta_1 + \theta_2)}$. **Example 1**: Express $z = -1 + i\sqrt{3}$ in polar and exponential form. Solution: $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$ $\theta = \tan^{-1}\left(\frac{\sqrt{3}}{-1}\right) = \tan^{-1}(-\sqrt{3})$. Since $z$ is in QII, $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ Polar: $z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$ Exponential: $z = 2e^{i2\pi/3}$ **Exercise 1**: Convert $z = 2 - 2i$ to polar and exponential form. Solution: $r = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2}$ $\theta = \tan^{-1}(-2/2) = \tan^{-1}(-1) = -\pi/4$ (QIV) Polar: $2\sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)$ Exponential: $2\sqrt{2} e^{-i\pi/4}$ **Exercise 2**: Find $|z|$ and $\arg(z)$ for $z = -3i$. Solution: $z = 0 - 3i$, so $r = \sqrt{0^2 + (-3)^2} = 3$ On negative imaginary axis: $\arg(z) = -\pi/2$ (principal value) Answer: $|z| = 3$, $\arg(z) = -\pi/2$ ## II. De Moivre's Theorem De Moivre’s theorem provides a powerful method for computing integer powers and roots of complex numbers in polar/exponential form. > **To retain**: For any integer $n$ and real $\theta$, $$ (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta) $$ Equivalently, $(e^{i\theta})^n = e^{in\theta}$. **Proof sketch (by induction)**: Base case $n=1$: trivial. Assume true for $n=k$: $(\cos\theta + i\sin\theta)^k = \cos(k\theta) + i\sin(k\theta)$ For $n=k+1$: LHS $= [\cos\theta + i\sin\theta]^{k+1} = [\cos\theta + i\sin\theta]^k \cdot (\cos\theta + i\sin\theta)$ $= [\cos(k\theta) + i\sin(k\theta)](\cos\theta + i\sin\theta)$ $= \cos(k\theta)\cos\theta - \sin(k\theta)\sin\theta + i[\sin(k\theta)\cos\theta + \cos(k\theta)\sin\theta]$ $= \cos((k+1)\theta) + i\sin((k+1)\theta)$ (using compound angle formulas) Hence true for all $n \in \mathbb{Z}^+$; extended to $n=0$ and negative $n$ via reciprocal. **Example 2**: Compute $(1 + i)^5$ using De Moivre’s theorem. Solution: First, $1+i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$ By De Moivre: $(1+i)^5 = (\sqrt{2})^5 \left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right) = 4\sqrt{2} \left(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = 4\sqrt{2} \cdot (-\frac{\sqrt{2}}{2})(1 + i) = -4(1+i) = -4 - 4i$ **Exercise 3**: Use De Moivre’s theorem to find $(-\sqrt{3} + i)^6$. Solution: $-\sqrt{3} + i$: $r = \sqrt{3 + 1} = 2$, $\theta = \tan^{-1}(1/(-\sqrt{3})) = \tan^{-1}(-1/\sqrt{3})$. QII → $\theta = \pi - \pi/6 = 5\pi/6$ So $z = 2(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6})$ $z^6 = 2^6 (\cos(6 \cdot \frac{5\pi}{6}) + i\sin(6 \cdot \frac{5\pi}{6})) = 64 (\cos 5\pi + i\sin 5\pi) = 64(-1 + i\cdot 0) = -64$ **Exercise 4**: Verify De Moivre’s theorem for $n=-2$ and $z = 2e^{i\pi/3}$. Solution: LHS: $(2e^{i\pi/3})^{-2} = 2^{-2} e^{-i2\pi/3} = \frac{1}{4} e^{-i2\pi/3}$ RHS: $2^{-2} e^{i(-2)(\pi/3)} = \frac{1}{4} e^{-i2\pi/3}$ → equal. ## III. Applications of De Moivre's Theorem De Moivre’s theorem is primarily used to: 1. Find $n$th powers of complex numbers 2. Find $n$th roots of complex numbers 3. Derive trigonometric identities ### Finding $n$th Roots The $n$ distinct $n$th roots of $z = re^{i\theta}$ are given by: $$ z_k = r^{1/n} \left[ \cos\left(\frac{\theta + 2k\pi}{n}\right) + i\sin\left(\frac{\theta + 2k\pi}{n}\right) \right], \quad k = 0, 1, \dots, n-1 $$ or equivalently $z_k = r^{1/n} e^{i(\theta + 2k\pi)/n}$ > **To retain**: The $n$th roots lie on a circle of radius $r^{1/n}$ and are equally spaced by angle $2\pi/n$. **Example 3**: Find the cube roots of $z = 8$. Solution: $8 = 8e^{i0}$ (principal argument 0) Cube roots: $z_k = 8^{1/3} e^{i(0 + 2k\pi)/3} = 2 e^{i2k\pi/3}$, $k=0,1,2$ - $k=0$: $2e^{i0} = 2$ - $k=1$: $2e^{i2\pi/3} = 2\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -1 + i\sqrt{3}$ - $k=2$: $2e^{i4\pi/3} = 2\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = -1 - i\sqrt{3}$ Roots: $2,\ -1 + i\sqrt{3},\ -1 - i\sqrt{3}$ **Exercise 5**: Find the fourth roots of $z = -16$. Solution: $-16 = 16e^{i\pi}$ Fourth roots: $z_k = 16^{1/4} e^{i(\pi + 2k\pi)/4} = 2 e^{i\pi(1+2k)/4}$, $k=0,1,2,3$ - $k=0$: $2e^{i\pi/4} = 2(\cos\pi/4 + i\sin\pi/4) = \sqrt{2} + i\sqrt{2}$ - $k=1$: $2e^{i3\pi/4} = -\sqrt{2} + i\sqrt{2}$ - $k=2$: $2e^{i5\pi/4} = -\sqrt{2} - i\sqrt{2}$ - $k=3$: $2e^{i7\pi/4} = \sqrt{2} - i\sqrt{2}$ Roots: $\pm\sqrt{2} \pm i\sqrt{2}$ (all combinations) **Exercise 6**: Solve $z^3 = 8i$. Solution: $8i = 8e^{i\pi/2}$ Cube roots: $z_k = 2 e^{i(\pi/2 + 2k\pi)/3} = 2 e^{i\pi(1+4k)/6}$, $k=0,1,2$ - $k=0$: $2e^{i\pi/6} = 2(\cos\pi/6 + i\sin\pi/6) = \sqrt{3} + i$ - $k=1$: $2e^{i5\pi/6} = 2(-\sqrt{3}/2 + i/2) = -\sqrt{3} + i$ - $k=2$: $2e^{i9\pi/6} = 2e^{i3\pi/2} = 2(0 - i) = -2i$ Roots: $\sqrt{3} + i,\ -\sqrt{3} + i,\ -2i$ ### Trigonometric Identities Using $z = \cos\theta + i\sin\theta$, we have $z^n + z^{-n} = 2\cos(n\theta)$ and $z^n - z^{-n} = 2i\sin(n\theta)$. **Example 4**: Express $\cos^3\theta$ in terms of cosines of multiple angles. Solution: Let $z = \cos\theta + i\sin\theta$, so $z^n = \cos n\theta + i\sin n\theta$ Then $z + z^{-1} = 2\cos\theta$ $\cos^3\theta = \left(\frac{z + z^{-1}}{2}\right)^3 = \frac{1}{8}(z^3 + 3z + 3z^{-1} + z^{-3}) = \frac{1}{8}[(z^3 + z^{-3}) + 3(z + z^{-1})]$ $= \frac{1}{8}[2\cos 3\theta + 3 \cdot 2\cos\theta] = \frac{1}{4}\cos 3\theta + \frac{3}{4}\cos\theta$ **Exercise 7**: Express $\sin^4\theta$ in terms of cosines of multiple angles. Solution: $\sin\theta = \frac{z - z^{-1}}{2i}$, so $\sin^4\theta = \left(\frac{z - z^{-1}}{2i}\right)^4 = \frac{(z - z^{-1})^4}{16}$ (since $i^4=1$) Expand: $(z - z^{-1})^4 = z^4 - 4z^2 + 6 - 4z^{-2} + z^{-4}$ So $\sin^4\theta = \frac{1}{16}(z^4 + z^{-4} - 4z^2 - 4z^{-2} + 6)$ $= \frac{1}{16}[2\cos 4\theta - 4\cdot 2\cos 2\theta + 6] = \frac{1}{8}\cos 4\theta - \frac{1}{2}\cos 2\theta + \frac{3}{8}$ ## IV. Loci in the Argand Diagram Loci are sets of points $z$ satisfying a given condition. We interpret equations and inequalities geometrically. ### Key Loci 1. **Circle**: $|z - z_0| = r$ → center $z_0$, radius $r$ 2. **Perpendicular bisector**: $|z - z_1| = |z - z_2|$ → line equidistant from $z_1, z_2$ 3. **Half-line (ray)**: $\arg(z - z_0) = \theta$ → ray from $z_0$ at angle $\theta$ 4. **Arc of circle**: $\arg\left(\frac{z - z_1}{z - z_2}\right) = \alpha$ → arc subtending angle $\alpha$ at $z_1, z_2$ (excluding endpoints) 5. **Line**: $\operatorname{Re}(z) = c$ → vertical line; $\operatorname{Im}(z) = c$ → horizontal line 6. **General line**: $\operatorname{Re}(\bar{a}z) = d$ or $|z - z_1| = k|z - z_2|$ (Circle of Apollonius if $k \neq 1$) > **To retain**: For $\arg(z - z_0) = \theta$, the locus is a half-line starting at $z_0$ (excluding $z_0$) making angle $\theta$ with positive real axis. **Example 5**: Sketch the locus of $|z - 2i| = 3$. Solution: This is a circle with center at $2i$ (i.e., $(0,2)$) and radius 3. **Exercise 8**: Describe the locus of $\arg(z + 1) = \pi/4$. Solution: $z + 1 = z - (-1)$, so $\arg(z - (-1)) = \pi/4$ → Half-line starting at $-1$ (on real axis) making angle $\pi/4$ with positive real axis. **Exercise 9**: Find the Cartesian equation of the locus $|z - 1| = |z + i|$. Solution: Let $z = x + iy$ $|(x-1) + iy| = |x + i(y+1)|$ $\sqrt{(x-1)^2 + y^2} = \sqrt{x^2 + (y+1)^2}$ Square both sides: $(x-1)^2 + y^2 = x^2 + (y+1)^2$ $x^2 - 2x + 1 + y^2 = x^2 + y^2 + 2y + 1$ Simplify: $-2x = 2y$ → $y = -x$ → Locus is the line $y = -x$ (through origin, slope -1) **Exercise 10**: Sketch $\left|\frac{z - 2}{z + 2}\right| = 1$. Solution: $|z-2| = |z+2|$ → set of points equidistant from $2$ and $-2$ on real axis → perpendicular bisector → imaginary axis ($x=0$) ## V. Elementary Transformations from the z-plane to the w-plane We consider mappings $w = f(z)$ where $z$ and $w$ are complex variables. These transform points, lines, and circles in the $z$-plane to corresponding sets in the $w$-plane. ### Common Transformations 1. **Translation**: $w = z + b$ → shifts every point by vector $b$ 2. **Rotation**: $w = e^{i\alpha} z$ → rotates about origin by angle $\alpha$ 3. **Enlargement (homothety)**: $w = az$ ($a \in \mathbb{R}^+$) → scales by factor $a$ about origin 4. **Combined**: $w = az + b$ → enlargement followed by translation 5. **Reciprocal**: $w = 1/z$ → maps lines/circles to lines/circles (with special cases) > **To retain**: The transformation $w = az + b$ ($a \neq 0$) is a similarity transformation: it preserves angles and maps lines to lines, circles to circles. **Example 6**: Find the image of the line $\operatorname{Im}(z) = 1$ under $w = (1+i)z$. Solution: Let $z = x + i$, so $w = (1+i)(x + i) = x + ix + i + i^2 = x + ix + i - 1 = (x-1) + i(x+1)$ Thus $w = u + iv$ where $u = x-1$, $v = x+1$ Eliminate $x$: from $u = x-1$ → $x = u+1$, substitute into $v$: $v = (u+1)+1 = u+2$ → Image is the line $v = u + 2$ in the $w$-plane. **Exercise 11**: Find the image of the circle $|z| = 2$ under $w = z + 3 - 2i$. Solution: Translation by $3 - 2i$: center moves from $0$ to $3 - 2i$, radius unchanged → $|w - (3 - 2i)| = 2$ **Exercise 12**: What is the image of the ray $\arg(z) = \pi/3$ under $w = 2z$? Solution: Enlargement by factor 2 about origin: scales distance but preserves direction → $\arg(w) = \arg(2z) = \arg(z) = \pi/3$ → Ray from origin at angle $\pi/3$ (unchanged direction, points twice as far) **Exercise 13**: Under $w = 1/z$, find the image of the line $\operatorname{Re}(z) = 1$. Solution: Let $z = x + iy$, $x=1$, so $z = 1 + iy$ $w = 1/(1 + iy) = (1 - iy)/(1 + y^2) = \frac{1}{1+y^2} - i\frac{y}{1+y^2}$ Let $w = u + iv$, so $u = \frac{1}{1+y^2}$, $v = -\frac{y}{1+y^2}$ Note $u > 0$. Also, $v^2 = \frac{y^2}{(1+y^2)^2} = \frac{1}{1+y^2} \cdot \frac{y^2}{1+y^2} = u(1 - u)$? Better: from $u = 1/(1+y^2)$ → $1+y^2 = 1/u$ → $y^2 = (1/u) - 1$ $v = -y/(1+y^2) = -y \cdot u$ → $y = -v/u$ Substitute: $(-v/u)^2 = (1/u) - 1$ → $v^2/u^2 = 1/u - 1$ → multiply by $u^2$: $v^2 = u - u^2$ → $u^2 - u + v^2 = 0$ → complete square: $(u - 1/2)^2 + v^2 = (1/2)^2$ → Circle center $(1/2, 0)$, radius $1/2$ in $w$-plane. Note: $z=0$ (on line? no, $x=1$ so $z\neq0$) maps to $w=\infty$; but line $\operatorname{Re}(z)=1$ does not pass through origin, so image is a circle not passing through origin. Actually, when $y\to\pm\infty$, $w\to 0$, so origin is on the circle? Check: $(0-1/2)^2 + 0^2 = 1/4$ → yes, origin lies on the circle. But $w=0$ corresponds to $z\to\infty$, which is on the extended line. So the image is the circle passing through origin: $|w - 1/2| = 1/2$. ## VI. Introduction to Plane Transformations and Matrix Representation Plane transformations can be represented using $2 \times 2$ real matrices acting on position vectors $\begin{pmatrix} x \\ y \end{pmatrix}$ corresponding to $z = x + iy$. A linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ is given by: $$ T\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix} $$ This corresponds to $w = (a+ic)z + (b+id)\bar{z}$? Not exactly — for **linear** transformations (fixing origin), we have $w = pz + q\bar{z}$ with $p,q \in \mathbb{C}$. But if we restrict to transformations expressible as $w = az$ ($a \in \mathbb{C}$), then it’s a similarity. However, general linear transformations (including shears, stretches) are represented by real $2\times2$ matrices. > **To retain**: The transformation $w = az$ with $a = re^{i\theta}$ corresponds to the matrix $$ \begin{pmatrix} r\cos\theta & -r\sin\theta \\ r\sin\theta & r\cos\theta \end{pmatrix} $$ which represents enlargement by $r$ and rotation by $\theta$. **Example 7**: Find the matrix for rotation by $\pi/6$ anticlockwise about origin. Solution: Rotation by $\theta$: $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ So for $\theta = \pi/6$: $\begin{pmatrix} \cos\pi/6 & -\sin\pi/6 \\ \sin\pi/6 & \cos\pi/6 \end{pmatrix} = \begin{pmatrix} \sqrt{3}/2 & -1/2 \\ 1/2 & \sqrt{3}/2 \end{pmatrix}$ **Exercise 14**: What is the matrix representation of enlargement by scale factor 3 about the origin? Solution: Enlargement by $k$: $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ → $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ **Exercise 15**: Find the matrix for reflection in the line $y = x$. Solution: Reflection in $y=x$ swaps $x$ and $y$: $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ ## VII. Reflection and Rotation Transformations ### Reflection A reflection is an isometry (distance-preserving) that flips the plane over a line (the mirror). - **Reflection in real axis ($\operatorname{Im}(z)=0$)**: $w = \bar{z}$ → matrix $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ - **Reflection in imaginary axis ($\operatorname{Re}(z)=0$)**: $w = -\bar{z}$ → matrix $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$ - **Reflection in line $y = x\tan\alpha$**: Formula: $w = e^{i2\alpha} \bar{z}$ Matrix: $\begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ \sin 2\alpha & -\cos 2\alpha \end{pmatrix}$ > **To retain**: Reflection in a line through origin making angle $\alpha$ with real axis: $w = e^{i2\alpha} \bar{z}$ **Example 8**: Find the reflection of $z = 2 + 3i$ in the line $y = \sqrt{3}x$ (i.e., $\alpha = \pi/3$). Solution: $\alpha = \pi/3$, so $2\alpha = 2\pi/3$ $w = e^{i2\pi/3} \overline{(2+3i)} = e^{i2\pi/3} (2 - 3i)$ $e^{i2\pi/3} = \cos 2\pi/3 + i\sin 2\pi/3 = -1/2 + i\sqrt{3}/2$ $w = (-1/2 + i\sqrt{3}/2)(2 - 3i) = (-1/2)(2) + (-1/2)(-3i) + (i\sqrt{3}/2)(2) + (i\sqrt{3}/2)(-3i)$ $= -1 + (3/2)i + i\sqrt{3} - (3i^2\sqrt{3}/2) = -1 + (3/2)i + i\sqrt{3} + (3\sqrt{3}/2)$ (since $i^2=-1$) Combine real: $-1 + 3\sqrt{3}/2$ Combine imag: $3/2 + \sqrt{3}$ So $w = \left(-1 + \frac{3\sqrt{3}}{2}\right) + i\left(\frac{3}{2} + \sqrt{3}\right)$ **Exercise 16**: Find the matrix for reflection in the line $y = 0$ (real axis). Solution: $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ **Exercise 17**: What is the image of $z = 1 - i$ under reflection in the line $y = x$? Solution: Reflection in $y=x$: $w = \bar{z}$ rotated? Actually, reflection in $y=x$ is $w = i \bar{z}$? Let’s derive: Line $y=x$ → $\alpha = \pi/4$, so $w = e^{i\pi/2} \bar{z} = i \bar{z}$ $\bar{z} = 1 + i$, so $w = i(1+i) = i + i^2 = i - 1 = -1 + i$ Alternatively, swap real and imag parts: $(x,y) \to (y,x)$ → $(1,-1) \to (-1,1)$ → $-1 + i$ Answer: $-1 + i$ ### Rotation A rotation about the origin by angle $\theta$ (anticlockwise) is given by: $$ w = e^{i\theta} z $$ Matrix: $\begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$ > **To retain**: Rotation preserves distances and orientation (it is a direct isometry). **Exercise 18**: Rotate $z = -1 + i$ by $\pi/2$ anticlockwise about origin. Solution: $w = e^{i\pi/2} z = i(-1 + i) = -i + i^2 = -i - 1 = -1 - i$ **Exercise 19**: Find the matrix for rotation by $-\pi/4$ (clockwise by $\pi/4$). Solution: $\theta = -\pi/4$ → $\begin{pmatrix} \cos(-\pi/4) & -\sin(-\pi/4) \\ \sin(-\pi/4) & \cos(-\pi/4) \end{pmatrix} = \begin{pmatrix} \sqrt{2}/2 & \sqrt{2}/2 \\ -\sqrt{2}/2 & \sqrt{2}/2 \end{pmatrix}$ ## VIII. Enlargement, Stretching, Shearing and Composite Transformations ### Enlargement (Homothety) $w = kz$, $k \in \mathbb{R}^+$ → scales distances from origin by factor $k$ Matrix: $\begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix}$ Preserves shape, changes size; orientation preserved if $k>0$. ### Stretching Non-uniform scaling: - Along real axis: $w = \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix} z$, $a \in \mathbb{R}$ - Along imaginary axis: $w = \begin{pmatrix} 1 & 0 \\ 0 & b \end{pmatrix} z$ Changes shape (e.g., circles become ellipses); orientation preserved if $a,b>0$. ### Shearing A shear shifts points parallel to a fixed line, proportional to their distance from that line. - **Shear parallel to x-axis**: $w = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} z$ → $(x,y) \mapsto (x + ky, y)$ - **Shear parallel to y-axis**: $w = \begin{pmatrix} 1 & 0 \\ k & 1 \end{pmatrix} z$ → $(x,y) \mapsto (x, y + kx)$ > **To retain**: Shears preserve area (determinant = 1) but do not preserve angles or shape. **Example 9**: Apply shear parallel to x-axis with $k=2$ to $z = 1 + i$. Solution: Matrix: $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1\cdot1 + 2\cdot1 \\ 0\cdot1 + 1\cdot1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$ → $w = 3 + i$ **Exercise 20**: Find the image of $z = 2 - 3i$ under a shear parallel to y-axis with $k = -1$. Solution: Matrix: $\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}$ $\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ -3 \end{pmatrix} = \begin{pmatrix} 1\cdot2 + 0\cdot(-3) \\ (-1)\cdot2 + 1\cdot(-3) \end{pmatrix} = \begin{pmatrix} 2 \\ -2 -3 \end{pmatrix} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}$ → $w = 2 - 5i$ **Exercise 21**: What is the matrix for stretching by factor 2 in the real direction and 1/2 in the imaginary direction? Solution: $\begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix}$ ### Composite Transformations When applying transformation $T_2$ after $T_1$, the combined matrix is $[T_2][T_1]$ (apply right to left). **Example 10**: Find the matrix for: first rotate by $\pi/3$, then enlarge by factor 2. Solution: Rotation $R = \begin{pmatrix} \cos\pi/3 & -\sin\pi/3 \\ \sin\pi/3 & \cos\pi/3 \end{pmatrix} = \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix}$ Enlargement $E = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$ Composite: $E \cdot R = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & 1/2 \end{pmatrix} = \begin{pmatrix} 2\cdot1/2 + 0\cdot\sqrt{3}/2 & 2\cdot(-\sqrt{3}/2) + 0\cdot1/2 \\ 0\cdot1/2 + 2\cdot\sqrt{3}/2 & 0\cdot(-\sqrt{3}/2) + 2\cdot1/2 \end{pmatrix} = \begin{pmatrix} 1 & -\sqrt{3} \\ \sqrt{3} & 1 \end{pmatrix}$ **Exercise 22**: Find the composite matrix for: reflection in real axis, then rotation by $\pi/4$. Solution: Reflection in real axis: $M_r = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ Rotation by $\pi/4$: $R = \begin{pmatrix} \cos\pi/4 & -\sin\pi/4 \\ \sin\pi/4 & \cos\pi/4 \end{pmatrix} = \begin{pmatrix} \sqrt{2}/2 & -\sqrt{2}/2 \\ \sqrt{2}/2 & \sqrt{2}/2 \end{pmatrix}$ Composite: $R \cdot M_r = \begin{pmatrix} \sqrt{2}/2 & -\sqrt{2}/2 \\ \sqrt{2}/2 & \sqrt{2}/2 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} \sqrt{2}/2\cdot1 + (-\sqrt{2}/2)\cdot0 & \sqrt{2}/2\cdot0 + (-\sqrt{2}/2)\cdot(-1) \\ \sqrt{2}/2\cdot1 + \sqrt{2}/2\cdot0 & \sqrt{2}/2\cdot0 + \sqrt{2}/2\cdot(-1) \end{pmatrix} = \begin{pmatrix} \sqrt{2}/2 & \sqrt{2}/2 \\ \sqrt{2}/2 & -\sqrt{2}/2 \end{pmatrix}$ ## IX. Introduction to Similitudes and Their Properties A similitude (or similarity transformation) is a transformation that preserves angles and maps figures to similar figures (same shape, possibly different size). It consists of an isometry (rotation, reflection, translation) followed by an enlargement. In complex form, a similitude is given by: $$ w = az + b \quad \text{where } a, b \in \mathbb{C}, \; a \neq 0 $$ - If $\operatorname{Im}(a) = 0$ and $a > 0$, it’s a direct similitude (orientation-preserving) - If $a$ is not real, it includes rotation - Translation $b$ shifts the result > **To retain**: The transformation $w = az + b$ ($a \neq 0$) is a similitude. It preserves angles and maps lines to lines, circles to circles. The linear part $az$ determines scaling and rotation: $|a|$ is the scale factor, $\arg(a)$ is the rotation angle. **Example 11**: Identify the similitude $w = (2i)z + (1 - i)$. Solution: $a = 2i = 2e^{i\pi/2}$ → scale factor $|a| = 2$, rotation $\arg(a) = \pi/2$ $b = 1 - i$ → translation by $1 - i$ So: enlarge by 2, rotate by $\pi/2$ anticlockwise, then translate by $1 - i$. **Exercise 23**: What is the scale factor and rotation angle of $w = (-3 + 3i)z$? Solution: $a = -3 + 3i = 3\sqrt{2} e^{i3\pi/4}$ (since modulus $= \sqrt{9+9}= \sqrt{18}=3\sqrt{2}$, argument $= \tan^{-1}(3/-3)=\tan^{-1}(-1)$ in QII → $3\pi/4$) Scale factor: $3\sqrt{2}$, rotation angle: $3\pi/4$ **Exercise 24**: Find the similitude that maps $z=0$ to $w=2+i$, $z=1$ to $w=4+2i$. Solution: Assume $w = az + b$ At $z=0$: $w = b = 2+i$ At $z=1$: $w = a(1) + b = a + (2+i) = 4+2i$ → $a = (4+2i) - (2+i) = 2 + i$ So $w = (2+i)z + (2+i)$