x < − 1 x \lt -1 x < − 1 x > 4 x \gt 4 x > 4
Example (HL): Solve x − 1 x + 2 ≤ 0 \frac{x - 1}{x + 2} \le 0 x + 2 x − 1 ≤ 0
Critical values: x = 1 x = 1 x = 1 x = − 2 x = -2 x = − 2
Sign chart:
Interval x < − 2 x < -2 x < − 2 − 2 < x < 1 -2 < x < 1 − 2 < x < 1 x > 1 x > 1 x > 1 x − 1 x - 1 x − 1 Negative Negative Positive x + 2 x + 2 x + 2 Negative Positive Positive Quotient Positive Negative Non-negative
Solution: x ∈ ( − 2 , 1 ] x \in (-2, 1] x ∈ ( − 2 , 1 ]
∣ x ∣ < a ⟺ − a < x < a |x| \lt a \iff -a \lt x \lt a ∣ x ∣ < a ⟺ − a < x < a
∣ x ∣ > a ⟺ x < − a |x| > a \iff x \lt -a ∣ x ∣ > a ⟺ x < − a x > a x > a x > a
Example: Solve ∣ 2 x − 3 ∣ < 5 |2x - 3| \lt 5 ∣2 x − 3∣ < 5
− 5 < 2 x − 3 < 5 -5 \lt 2x - 3 \lt 5 − 5 < 2 x − 3 < 5
− 2 < 2 x < 8 -2 \lt 2x \lt 8 − 2 < 2 x < 8
− 1 < x < 4 -1 \lt x \lt 4 − 1 < x < 4
A complex number is of the form z = a + b i z = a + bi z = a + bi a , b ∈ R a, b \in \mathbb{R} a , b ∈ R i 2 = − 1 i^2 = -1 i 2 = − 1
a a a real part (Re ( z ) \operatorname{Re}(z) Re ( z ) b b b imaginary part (Im ( z ) \operatorname{Im}(z) Im ( z ) Addition: ( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i (a + bi) + (c + di) = (a + c) + (b + d)i ( a + bi ) + ( c + d i ) = ( a + c ) + ( b + d ) i
Multiplication: ( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i (a + bi)(c + di) = (ac - bd) + (ad + bc)i ( a + bi ) ( c + d i ) = ( a c − b d ) + ( a d + b c ) i
Complex conjugate: z ˉ = a − b i \bar{z} = a - bi z ˉ = a − bi
Z ⋅ z ˉ = a 2 + b 2 Z \cdot \bar{z} = a^2 + b^2 Z ⋅ z ˉ = a 2 + b 2 Division: a + b i c + d i = ( a + b i ) ( c − d i ) ( c + d i ) ( c − d i ) = ( a c + b d ) + ( b c − a d ) i c 2 + d 2 \dfrac{a + bi}{c + di} = \dfrac{(a + bi)(c - di)}{(c + di)(c - di)} = \dfrac{(ac + bd) + (bc - ad)i}{c^2 + d^2} c + d i a + bi = ( c + d i ) ( c − d i ) ( a + bi ) ( c − d i ) = c 2 + d 2 ( a c + b d ) + ( b c − a d ) i
The modulus (or absolute value) of z = a + b i z = a + bi z = a + bi
∣ z ∣ = a 2 + b 2 |z| = \sqrt{a^2 + b^2} ∣ z ∣ = a 2 + b 2 The argument arg ( z ) \arg(z) arg ( z ) θ \theta θ
θ = arctan ( b a ) , a > 0 \theta = \arctan\left(\frac{b}{a}\right), \quad a > 0 θ = arctan ( a b ) , a > 0 For a < 0 a < 0 a < 0 π \pi π
Quadrant check for the argument:
| Quadrant | a a a b b b arg ( z ) \arg(z) arg ( z ) > 0 > 0 > 0 > 0 > 0 > 0 arctan ( b / a ) \arctan(b/a) arctan ( b / a ) < 0 < 0 < 0 > 0 > 0 > 0 π − arctan ( b / ∣ a ∣ ) \pi - \arctan(b/ | a | ) π − arctan ( b /∣ a ∣ ) < 0 < 0 < 0 < 0 < 0 < 0 − π + arctan ( b / a ) -\pi + \arctan(b/a) − π + arctan ( b / a ) > 0 > 0 > 0 < 0 < 0 < 0 − arctan ( b / a ) -\arctan(b/a) − arctan ( b / a )
Z = r ( cos θ + i sin θ ) = r e i θ Z = r(\cos\theta + i\sin\theta) = re^{i\theta} Z = r ( cos θ + i sin θ ) = r e i θ Where r = ∣ z ∣ r = |z| r = ∣ z ∣ θ = arg ( z ) \theta = \arg(z) θ = arg ( z )
Example (HL): Express z = − 1 + i 3 z = -1 + i\sqrt{3} z = − 1 + i 3
r = 1 + 3 = 2 r = \sqrt{1 + 3} = 2 r = 1 + 3 = 2 a = − 1 < 0 a = -1 < 0 a = − 1 < 0 b = 3 > 0 b = \sqrt{3} > 0 b = 3 > 0
θ = π − arctan ( 3 1 ) = π − π 3 = 2 π 3 \theta = \pi - \arctan\!\left(\frac{\sqrt{3}}{1}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3} θ = π − arctan ( 1 3 ) = π − 3 π = 3 2 π Z = 2 ( cos 2 π 3 + i sin 2 π 3 ) = 2 e 2 π i / 3 Z = 2\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 2e^{2\pi i/3} Z = 2 ( cos 3 2 π + i sin 3 2 π ) = 2 e 2 π i /3 For z = r ( cos θ + i sin θ ) z = r(\cos\theta + i\sin\theta) z = r ( cos θ + i sin θ ) n ∈ Z n \in \mathbb{Z} n ∈ Z
Z n = r n ( cos n θ + i sin n θ ) Z^n = r^n(\cos n\theta + i\sin n\theta) Z n = r n ( cos n θ + i sin n θ ) Proof by induction for positive integers. Base case n = 1 n = 1 n = 1 n = k n = k n = k z k + 1 = z k ⋅ z = r k ( cos k θ + i sin k θ ) ⋅ r ( cos θ + i sin θ ) z^{k+1} = z^k \cdot z = r^k(\cos k\theta + i\sin k\theta) \cdot r(\cos\theta + i\sin\theta) z k + 1 = z k ⋅ z = r k ( cos k θ + i sin k θ ) ⋅ r ( cos θ + i sin θ ) r k + 1 ( cos ( k + 1 ) θ + i sin ( k + 1 ) θ ) r^{k+1}(\cos(k+1)\theta + i\sin(k+1)\theta) r k + 1 ( cos ( k + 1 ) θ + i sin ( k + 1 ) θ )
Example: Express ( 3 + i ) 4 (\sqrt{3} + i)^4 ( 3 + i ) 4 a + b i a + bi a + bi
First find modulus and argument: r = 2 r = 2 r = 2 θ = π 6 \theta = \frac{\pi}{6} θ = 6 π
( 3 + i ) 4 = 2 4 ( cos 4 π 6 + i sin 4 π 6 ) = 16 ( cos 2 π 3 + i sin 2 π 3 ) = 16 ( − 1 2 + i 3 2 ) = − 8 + 8 3 i (\sqrt{3} + i)^4 = 2^4\left(\cos\frac{4\pi}{6} + i\sin\frac{4\pi}{6}\right) = 16\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right) = 16\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -8 + 8\sqrt{3}\,i ( 3 + i ) 4 = 2 4 ( cos 6 4 π + i sin 6 4 π ) = 16 ( cos 3 2 π + i sin 3 2 π ) = 16 ( − 2 1 + i 2 3 ) = − 8 + 8 3 i Example (HL): Express z = 1 − i 3 z = 1 - i\sqrt{3} z = 1 − i 3 z 5 z^5 z 5
r = 1 + 3 = 2 r = \sqrt{1 + 3} = 2 r = 1 + 3 = 2 a = 1 > 0 a = 1 > 0 a = 1 > 0 b = − 3 < 0 b = -\sqrt{3} < 0 b = − 3 < 0
θ = − π 3 , z = 2 e − π i / 3 \theta = -\frac{\pi}{3}, \quad z = 2e^{-\pi i/3} θ = − 3 π , z = 2 e − π i /3 Z 5 = 2 5 e − 5 π i / 3 = 32 ( cos − 5 π 3 + i sin − 5 π 3 ) = 32 ( 1 2 − i 3 2 ) = 16 − 16 3 i Z^5 = 2^5 e^{-5\pi i/3} = 32\left(\cos\frac{-5\pi}{3} + i\sin\frac{-5\pi}{3}\right) = 32\left(\frac{1}{2} - i\frac{\sqrt{3}}{2}\right) = 16 - 16\sqrt{3}\,i Z 5 = 2 5 e − 5 π i /3 = 32 ( cos 3 − 5 π + i sin 3 − 5 π ) = 32 ( 2 1 − i 2 3 ) = 16 − 16 3 i The n n n z n = 1 z^n = 1 z n = 1
Z k = cos 2 k π n + i sin 2 k π n , k = 0 , 1 , 2 , … , n − 1 Z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \quad k = 0, 1, 2, \ldots, n-1 Z k = cos n 2 k π + i sin n 2 k π , k = 0 , 1 , 2 , … , n − 1 These lie on the unit circle in the complex plane, equally spaced at angles of 2 π n \frac{2\pi}{n} n 2 π
Properties:
The sum of all n n n 0 0 0 The product of all n n n ( − 1 ) n − 1 (-1)^{n-1} ( − 1 ) n − 1 The n n n w = r e i θ w = re^{i\theta} w = r e i θ r n e i ( θ + 2 k π ) / n \sqrt[n]{r}\, e^{i(\theta + 2k\pi)/n} n r e i ( θ + 2 k π ) / n k = 0 , 1 , … , n − 1 k = 0, 1, \ldots, n-1 k = 0 , 1 , … , n − 1 Example: Find the cube roots of unity.
For z 3 = 1 z^3 = 1 z 3 = 1 z k = cos 2 k π 3 + i sin 2 k π 3 z_k = \cos\frac{2k\pi}{3} + i\sin\frac{2k\pi}{3} z k = cos 3 2 k π + i sin 3 2 k π k = 0 , 1 , 2 k = 0, 1, 2 k = 0 , 1 , 2
Z 0 = 1 , z 1 = − 1 2 + i 3 2 , z 2 = − 1 2 − i 3 2 Z_0 = 1, \quad z_1 = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \quad z_2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} Z 0 = 1 , z 1 = − 2 1 + i 2 3 , z 2 = − 2 1 − i 2 3 Note: 1 + z 1 + z 2 = 0 1 + z_1 + z_2 = 0 1 + z 1 + z 2 = 0
Example (HL): Find all complex numbers z z z z 4 = 16 z^4 = 16 z 4 = 16
The four fourth roots of 16 = 16 e 0 i 16 = 16e^{0i} 16 = 16 e 0 i
Z k = 2 e 2 k π i / 4 , k = 0 , 1 , 2 , 3 Z_k = 2\,e^{2k\pi i/4}, \quad k = 0, 1, 2, 3 Z k = 2 e 2 k π i /4 , k = 0 , 1 , 2 , 3 Z 0 = 2 , z 1 = 2 i , z 2 = − 2 , z 3 = − 2 i Z_0 = 2, \quad z_1 = 2i, \quad z_2 = -2, \quad z_3 = -2i Z 0 = 2 , z 1 = 2 i , z 2 = − 2 , z 3 = − 2 i A matrix A A A m × n m \times n m × n m m m n n n
Addition: Add corresponding elements (matrices must be the same order).
Scalar multiplication: Multiply every element by the scalar.
Matrix multiplication: If A A A m × p m \times p m × p B B B p × n p \times n p × n A B AB A B m × n m \times n m × n
The ( i , j ) (i, j) ( i , j ) A B AB A B
( A B ) i j = ∑ k = 1 p a i k b k j (AB)_{ij} = \sum_{k=1}^{p} a_{ik}b_{kj} ( A B ) ij = k = 1 ∑ p a ik b k j For A = ( a b c d ) A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} A = ( a c b d )
det ( A ) = a d − b c \det(A) = ad - bc det ( A ) = a d − b c If det ( A ) ≠ 0 \det(A) \neq 0 det ( A ) = 0
A − 1 = 1 a d − b c ( d − b − c a ) A^{-1} = \frac{1}{ad - bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix} A − 1 = a d − b c 1 ( d − c − b a ) For A = ( a b c d e f g h k ) A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & k \end{pmatrix} A = a d g b e h c f k
det ( A ) = a ( e k − f h ) − b ( d k − f g ) + c ( d h − e g ) \det(A) = a(ek - fh) - b(dk - fg) + c(dh - eg) det ( A ) = a ( e k − f h ) − b ( d k − f g ) + c ( d h − e g ) For A X = B AX = B A X = B A A A
X = A − 1 B X = A^{-1}B X = A − 1 B Example: Solve:
{ 2 x + y = 5 X − y = 1 \begin{cases} 2x + y = 5 \\ X - y = 1 \end{cases} { 2 x + y = 5 X − y = 1 ( 2 1 1 − 1 ) ( x y ) = ( 5 1 ) \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 1 \end{pmatrix} ( 2 1 1 − 1 ) ( x y ) = ( 5 1 ) det ( A ) = − 2 − 1 = − 3 \det(A) = -2 - 1 = -3 det ( A ) = − 2 − 1 = − 3 A − 1 = 1 − 3 ( − 1 − 1 − 1 2 ) = ( 1 3 1 3 1 3 − 2 3 ) A^{-1} = \frac{1}{-3}\begin{pmatrix} -1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{pmatrix} A − 1 = − 3 1 ( − 1 − 1 − 1 2 ) = ( 3 1 3 1 3 1 − 3 2 ) ( x y ) = ( 1 3 1 3 1 3 − 2 3 ) ( 5 1 ) = ( 2 1 ) \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{2}{3} \end{pmatrix}\begin{pmatrix} 5 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix} ( x y ) = ( 3 1 3 1 3 1 − 3 2 ) ( 5 1 ) = ( 2 1 ) Base case: Show the statement holds for n = 1 n = 1 n = 1 Inductive hypothesis: Assume the statement holds for n = k n = k n = k Inductive step: Show that if it holds for n = k n = k n = k n = k + 1 n = k + 1 n = k + 1 Conclusion: By the principle of mathematical induction, the statement holds for all n ≥ 1 n \geq 1 n ≥ 1 Why induction works. The base case anchors the chain at n = 1 n = 1 n = 1
Example: Prove by induction that ∑ r = 1 n r 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} ∑ r = 1 n r 2 = 6 n ( n + 1 ) ( 2 n + 1 )
Base case (n = 1 n = 1 n = 1 = 1 = 1 = 1 = 1 ⋅ 2 ⋅ 3 6 = 1 = \frac{1 \cdot 2 \cdot 3}{6} = 1 = 6 1 ⋅ 2 ⋅ 3 = 1
Inductive hypothesis: Assume ∑ r = 1 k r 2 = k ( k + 1 ) ( 2 k + 1 ) 6 \sum_{r=1}^{k} r^2 = \frac{k(k+1)(2k+1)}{6} ∑ r = 1 k r 2 = 6 k ( k + 1 ) ( 2 k + 1 )
Inductive step:
∑ r = 1 k + 1 r 2 = k ( k + 1 ) ( 2 k + 1 ) 6 + ( k + 1 ) 2 \sum_{r=1}^{k+1} r^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2 r = 1 ∑ k + 1 r 2 = 6 k ( k + 1 ) ( 2 k + 1 ) + ( k + 1 ) 2 = k ( k + 1 ) ( 2 k + 1 ) + 6 ( k + 1 ) 2 6 = \frac{k(k+1)(2k+1) + 6(k+1)^2}{6} = 6 k ( k + 1 ) ( 2 k + 1 ) + 6 ( k + 1 ) 2 = ( k + 1 ) [ k ( 2 k + 1 ) + 6 ( k + 1 ) ] 6 = \frac{(k+1)[k(2k+1) + 6(k+1)]}{6} = 6 ( k + 1 ) [ k ( 2 k + 1 ) + 6 ( k + 1 )] = ( k + 1 ) ( 2 k 2 + 7 k + 6 ) 6 = \frac{(k+1)(2k^2 + 7k + 6)}{6} = 6 ( k + 1 ) ( 2 k 2 + 7 k + 6 ) = ( k + 1 ) ( k + 2 ) ( 2 k + 3 ) 6 = \frac{(k+1)(k+2)(2k+3)}{6} = 6 ( k + 1 ) ( k + 2 ) ( 2 k + 3 ) = ( k + 1 ) ( ( k + 1 ) + 1 ) ( 2 ( k + 1 ) + 1 ) 6 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6} = 6 ( k + 1 ) (( k + 1 ) + 1 ) ( 2 ( k + 1 ) + 1 ) This matches the formula with n = k + 1 n = k + 1 n = k + 1 n ≥ 1 n \ge 1 n ≥ 1
Example: Prove that ∑ r = 1 n r 3 = n 2 ( n + 1 ) 2 4 \sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4} ∑ r = 1 n r 3 = 4 n 2 ( n + 1 ) 2
Base case (n = 1 n = 1 n = 1 = 1 = 1 = 1 = 1 ⋅ 4 4 = 1 = \frac{1 \cdot 4}{4} = 1 = 4 1 ⋅ 4 = 1
Inductive hypothesis: Assume ∑ r = 1 k r 3 = k 2 ( k + 1 ) 2 4 \sum_{r=1}^{k} r^3 = \frac{k^2(k+1)^2}{4} ∑ r = 1 k r 3 = 4 k 2 ( k + 1 ) 2
Inductive step:
∑ r = 1 k + 1 r 3 = k 2 ( k + 1 ) 2 4 + ( k + 1 ) 3 \sum_{r=1}^{k+1} r^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3 r = 1 ∑ k + 1 r 3 = 4 k 2 ( k + 1 ) 2 + ( k + 1 ) 3 = k 2 ( k + 1 ) 2 + 4 ( k + 1 ) 3 4 = \frac{k^2(k+1)^2 + 4(k+1)^3}{4} = 4 k 2 ( k + 1 ) 2 + 4 ( k + 1 ) 3 = ( k + 1 ) 2 [ k 2 + 4 ( k + 1 ) ] 4 = \frac{(k+1)^2[k^2 + 4(k+1)]}{4} = 4 ( k + 1 ) 2 [ k 2 + 4 ( k + 1 )] = ( k + 1 ) 2 ( k 2 + 4 k + 4 ) 4 = \frac{(k+1)^2(k^2 + 4k + 4)}{4} = 4 ( k + 1 ) 2 ( k 2 + 4 k + 4 ) = ( k + 1 ) 2 ( k + 2 ) 2 4 = \frac{(k+1)^2(k+2)^2}{4} = 4 ( k + 1 ) 2 ( k + 2 ) 2 = ( k + 1 ) 2 ( ( k + 1 ) + 1 ) 2 4 = \frac{(k+1)^2((k+1)+1)^2}{4} = 4 ( k + 1 ) 2 (( k + 1 ) + 1 ) 2 By induction, the result holds for all n ≥ 1 n \ge 1 n ≥ 1
To divide P ( x ) P(x) P ( x ) ( x − a ) (x - a) ( x − a )
P ( x ) = ( x − a ) Q ( x ) + R P(x) = (x - a)Q(x) + R P ( x ) = ( x − a ) Q ( x ) + R Where Q ( x ) Q(x) Q ( x ) R R R Remainder Theorem , R = P ( a ) R = P(a) R = P ( a )
Factor Theorem: ( x − a ) (x - a) ( x − a ) P ( x ) P(x) P ( x ) P ( a ) = 0 P(a) = 0 P ( a ) = 0
Example: Factorise x 3 − 3 x + 2 x^3 - 3x + 2 x 3 − 3 x + 2
Try P ( 1 ) = 1 − 3 + 2 = 0 P(1) = 1 - 3 + 2 = 0 P ( 1 ) = 1 − 3 + 2 = 0 ( x − 1 ) (x - 1) ( x − 1 )
Dividing: x 3 − 3 x + 2 = ( x − 1 ) ( x 2 + x − 2 ) = ( x − 1 ) ( x + 2 ) ( x − 1 ) = ( x − 1 ) 2 ( x + 2 ) x^3 - 3x + 2 = (x - 1)(x^2 + x - 2) = (x - 1)(x + 2)(x - 1) = (x - 1)^2(x + 2) x 3 − 3 x + 2 = ( x − 1 ) ( x 2 + x − 2 ) = ( x − 1 ) ( x + 2 ) ( x − 1 ) = ( x − 1 ) 2 ( x + 2 )
Example (HL): When P ( x ) = x 3 + 2 x 2 − 5 x − 6 P(x) = x^3 + 2x^2 - 5x - 6 P ( x ) = x 3 + 2 x 2 − 5 x − 6 ( x − 1 ) (x - 1) ( x − 1 ) P ( 1 ) = 1 + 2 − 5 − 6 = − 8 P(1) = 1 + 2 - 5 - 6 = -8 P ( 1 ) = 1 + 2 − 5 − 6 = − 8
Example: Solve ( x − 1 ) 2 ( x + 2 ) > 0 (x - 1)^2(x + 2) \gt 0 ( x − 1 ) 2 ( x + 2 ) > 0
The critical values are x = − 2 x = -2 x = − 2 x = 1 x = 1 x = 1
Sign chart:
Interval x < − 2 x < -2 x < − 2 − 2 < x < 1 -2 < x < 1 − 2 < x < 1 x > 1 x > 1 x > 1 ( x + 2 ) (x+2) ( x + 2 ) Negative Positive Positive ( x − 1 ) 2 (x-1)^2 ( x − 1 ) 2 Positive Positive Positive Product Negative Positive Positive
Solution: x < − 2 x < -2 x < − 2 x > 1 x > 1 x > 1 x ∈ ( − ∞ , − 2 ) ∪ ( 1 , ∞ ) x \in (-\infty, -2) \cup (1, \infty) x ∈ ( − ∞ , − 2 ) ∪ ( 1 , ∞ )
Note that x = − 1 x = -1 x = − 1 x = 1 x = 1 x = 1
Algebra is the mathematics of the unknown — it lets you solve for x when you do not know what x is, which is the foundation of all quantitative reasoning. Equations are balance scales: whatever you do to one side, you must do to the other. Inequalities are like equations with a range of solutions rather than a single answer. Complex numbers extend the number line into a plane, opening up solutions to equations that have no real answers. Matrices are grids of numbers that represent transformations — they can rotate, scale, and shear shapes in ways that are essential for computer graphics and physics simulations.
See the examples integrated throughout the sections above.
Sign errors in factorisation and expansion — always double-check by expanding back.Forgetting to reverse the inequality when multiplying/dividing by a negative number.Confusing the discriminant conditions for real vs. Complex roots. Δ < 0 \Delta < 0 Δ < 0 De Moivre’s theorem requires the argument to be in radians.Matrix multiplication is not commutative: A B ≠ B A AB \neq BA A B = B A Induction base case — always state and verify it explicitly. A proof without a base case is like a chain with no anchor.Domain restrictions on logarithms — log a ( x ) \log_a(x) log a ( x ) x > 0 x > 0 x > 0 Forgetting absolute values in the quadratic formula when Δ < 0 \Delta < 0 Δ < 0 x = − b ± i ∣ Δ ∣ 2 a x = \frac{-b \pm i\sqrt{|\Delta|}}{2a} x = 2 a − b ± i ∣Δ∣ Confusing the argument quadrant. For z = − 1 + i z = -1 + i z = − 1 + i arg ( z ) = 3 π 4 \arg(z) = \frac{3\pi}{4} arg ( z ) = 4 3 π arctan ( − 1 ) \arctan(-1) arctan ( − 1 ) Expand ( 3 x − 2 ) ( 2 x + 5 ) (3x - 2)(2x + 5) ( 3 x − 2 ) ( 2 x + 5 ) Factorise x 2 − 7 x + 12 x^2 - 7x + 12 x 2 − 7 x + 12 Solve 4 x − 7 = 2 x + 9 4x - 7 = 2x + 9 4 x − 7 = 2 x + 9 Solve x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 Solve 2 x 2 + x − 3 = 0 2x^2 + x - 3 = 0 2 x 2 + x − 3 = 0 Prove by induction that ∑ r = 1 n r 3 = n 2 ( n + 1 ) 2 4 \sum_{r=1}^{n} r^3 = \frac{n^2(n+1)^2}{4} ∑ r = 1 n r 3 = 4 n 2 ( n + 1 ) 2 Express z = 1 − i 3 z = 1 - i\sqrt{3} z = 1 − i 3 z 5 z^5 z 5 Find the modulus and argument of 1 + i 1 − i \frac{1 + i}{1 - i} 1 − i 1 + i Find the values of k k k k x 2 + 4 x + k = 0 kx^2 + 4x + k = 0 k x 2 + 4 x + k = 0 Given A = ( 1 2 3 4 ) A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} A = ( 1 3 2 4 ) B = ( 0 − 1 2 3 ) B = \begin{pmatrix} 0 & -1 \\ 2 & 3 \end{pmatrix} B = ( 0 2 − 1 3 ) A B − B A AB - BA A B − B A Find all complex numbers z z z z 4 = 16 z^4 = 16 z 4 = 16 Solve the inequality x 2 − 2 x − 15 > 0 x^2 - 2x - 15 \gt 0 x 2 − 2 x − 15 > 0 Factorise x 4 − 1 x^4 - 1 x 4 − 1 Find the remainder when P ( x ) = 2 x 3 − 3 x 2 + 5 x − 7 P(x) = 2x^3 - 3x^2 + 5x - 7 P ( x ) = 2 x 3 − 3 x 2 + 5 x − 7 ( x + 2 ) (x + 2) ( x + 2 ) Solve ∣ 3 x − 1 ∣ ≤ 8 |3x - 1| \le 8 ∣3 x − 1∣ ≤ 8 Find the modulus and argument of z = 3 + 4 i 1 − 2 i z = \frac{3 + 4i}{1 - 2i} z = 1 − 2 i 3 + 4 i Prove by induction that 3 n ≥ 2 n + n 3^n \ge 2^n + n 3 n ≥ 2 n + n n ≥ 1 n \ge 1 n ≥ 1 Express 2 x + 1 ( x + 1 ) ( x − 2 ) \frac{2x + 1}{(x+1)(x-2)} ( x + 1 ) ( x − 2 ) 2 x + 1 Solve z 3 = − 27 z^3 = -27 z 3 = − 27 Find the quadratic equation whose roots are 2 + 3 2 + \sqrt{3} 2 + 3 2 − 3 2 - \sqrt{3} 2 − 3 Given z = 2 + 3 i z = 2 + 3i z = 2 + 3 i w = 1 − 4 i w = 1 - 4i w = 1 − 4 i z w ˉ z\bar{w} z w ˉ ∣ z / w ∣ |z/w| ∣ z / w ∣ Solve the simultaneous equations x + i y + i z = 0 x + iy + iz = 0 x + i y + i z = 0 x − 2 y + i z = 1 + i x - 2y + iz = 1 + i x − 2 y + i z = 1 + i x x x y y y Prove by induction that 1 1 × 2 + 1 2 × 3 + ⋯ + 1 n ( n + 1 ) = n n + 1 \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} 1 × 2 1 + 2 × 3 1 + ⋯ + n ( n + 1 ) 1 = n + 1 n Find the matrix A A A A ( 1 2 ) = ( 5 3 ) A\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix} A ( 1 2 ) = ( 5 3 ) A ( 3 1 ) = ( 7 7 ) A\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} 7 \\ 7 \end{pmatrix} A ( 3 1 ) = ( 7 7 ) Express 3 x 2 − x + 2 ( x − 1 ) ( x 2 + 1 ) \frac{3x^2 - x + 2}{(x-1)(x^2 + 1)} ( x − 1 ) ( x 2 + 1 ) 3 x 2 − x + 2 Find all complex numbers z z z ∣ z − 2 i ∣ = ∣ z + 2 ∣ |z - 2i| = |z + 2| ∣ z − 2 i ∣ = ∣ z + 2∣ Prove that if a quadratic equation with rational coefficients has one irrational root a + b c a + b\sqrt{c} a + b c b ≠ 0 b \neq 0 b = 0 a − b c a - b\sqrt{c} a − b c A[1_Algebra] --> B[Key Concepts]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
D --> G[Real-world usage]
This topic covers the mathematical techniques and concepts related to algebra, including key theorems, methods, and problem-solving approaches.
Key concepts include:
quadratic equations and the discriminant simultaneous equations polynomial division and the factor theorem partial fractions binomial expansion Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.
