FAC1004 — Must-Memorize Cheat Sheet

Everything you need cold. No derivations. No explanations. Just the formulas and procedures that must be instant.


Complex Numbers

Forms

Form Expression
Cartesian $z = a + bi$
Polar $z = r(\cos\theta + i\sin\theta)$
Exponential $z = re^{i\theta}$

Conversions

$r = \sqrt{a^2 + b^2}$, $\theta = \tan^{-1}(b/a)$ (check quadrant) $a = r\cos\theta$, $b = r\sin\theta$ Principal argument: $-\pi < \theta \leq \pi$

De Moivre

$$z^n = r^n(\cos n\theta + i\sin n\theta) = r^n e^{in\theta}$$ $$z^{1/n} = \sqrt[n]{r}\left[\cos\frac{\theta + 2\pi k}{n} + i\sin\frac{\theta + 2\pi k}{n}\right], \quad k = 0,1,\ldots,n-1$$

Loci

  • $|z - z_0| = r$ → Circle (centre $z_0$, radius $r$) → $(x-a)^2 + (y-b)^2 = r^2$
  • $|z - z_1| = |z - z_2|$ → Perpendicular bisector of segment joining $z_1$, $z_2$
  • $\arg(z - z_0) = \theta$ → Ray (half-line) from $z_0$ at angle $\theta$, $z_0$ excluded
  • $<$ or $>$ → dotted boundary; $\leq$ or $\geq$ → solid boundary

Key Identities

$$z + \frac{1}{z} = 2\cos x, \quad z - \frac{1}{z} = 2i\sin x$$ $$z^n + z^{-n} = 2\cos(nx), \quad z^n - z^{-n} = 2i\sin(nx)$$

Complex Exponential

$$e^{a+bi} = e^a(\cos b + i\sin b)$$ $$|e^{a+bi}| = e^a, \quad \arg(e^{a+bi}) = b$$

Complex Logarithm

$$\ln(z) = \ln|z| + i(\arg(z) + 2\pi k), \quad k \in \mathbb{Z}$$ $$\text{Ln}(z) = \ln|z| + i\arg(z) \quad (\text{principal})$$

Complex Sine & Cosine

$$\cos z = \frac{e^{zi} + e^{-zi}}{2}, \qquad \sin z = \frac{e^{zi} - e^{-zi}}{2i}$$

Impedance (Application)

$$Z = R + i\left(\omega L - \frac{1}{\omega C}\right), \qquad V = IZ$$


Inverse Trigonometric Functions

Domains & Ranges

Function Domain Range (Principal)
$\sin^{-1} x$ $[-1, 1]$ $[-\pi/2, \pi/2]$
$\cos^{-1} x$ $[-1, 1]$ $[0, \pi]$
$\tan^{-1} x$ $\mathbb{R}$ $(-\pi/2, \pi/2)$
$\cot^{-1} x$ $\mathbb{R}$ $(0, \pi)$
$\sec^{-1} x$ $ x
$\csc^{-1} x$ $ x

Key Identities

$$\sin^{-1}(-x) = -\sin^{-1} x, \quad \tan^{-1}(-x) = -\tan^{-1} x$$ $$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}, \quad \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$$ $$\tan^{-1} x \pm \tan^{-1} y = \tan^{-1}\left(\frac{x \pm y}{1 \mp xy}\right)$$ $$2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$$

Interconversion

$$\sin^{-1} x = \cos^{-1}\sqrt{1-x^2} = \tan^{-1}\frac{x}{\sqrt{1-x^2}}$$

Derivatives

$$\frac{d}{dx}\sin^{-1}u = \frac{1}{\sqrt{1-u^2}}u', \quad \frac{d}{dx}\cos^{-1}u = \frac{-1}{\sqrt{1-u^2}}u'$$ $$\frac{d}{dx}\tan^{-1}u = \frac{1}{1+u^2}u', \quad \frac{d}{dx}\cot^{-1}u = \frac{-1}{1+u^2}u'$$ $$\frac{d}{dx}\sec^{-1}u = \frac{u'}{|u|\sqrt{u^2-1}}, \quad \frac{d}{dx}\csc^{-1}u = \frac{-u'}{|u|\sqrt{u^2-1}}$$


Hyperbolic Functions

Definitions

$$\sinh x = \frac{e^x - e^{-x}}{2}, \quad \cosh x = \frac{e^x + e^{-x}}{2}, \quad \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$$

Key Identity

$$\cosh^2 x - \sinh^2 x = 1$$ $$1 - \tanh^2 x = \text{sech}^2 x, \qquad \coth^2 x - 1 = \text{csch}^2 x$$

Addition & Double Angle

$$\sinh(x\pm y) = \sinh x\cosh y \pm \cosh x\sinh y$$ $$\cosh(x\pm y) = \cosh x\cosh y \pm \sinh x\sinh y$$ $$\sinh 2x = 2\sinh x\cosh x, \qquad \cosh 2x = \cosh^2 x + \sinh^2 x$$

Derivatives

$$\frac{d}{dx}\sinh u = \cosh u \cdot u', \quad \frac{d}{dx}\cosh u = \sinh u \cdot u'$$ $$\frac{d}{dx}\tanh u = \text{sech}^2 u \cdot u', \quad \frac{d}{dx}\coth u = -\text{csch}^2 u \cdot u'$$

Integrals

$$\int \sinh u , du = \cosh u + C, \quad \int \cosh u , du = \sinh u + C$$ $$\int \text{sech}^2 u , du = \tanh u + C, \quad \int \tanh x , dx = \ln(\cosh x) + C$$


Inverse Hyperbolic Functions

Logarithmic Forms

$$\sinh^{-1} x = \ln(x + \sqrt{x^2+1})$$ $$\cosh^{-1} x = \ln(x + \sqrt{x^2-1})$$ $$\tanh^{-1} x = \frac12\ln\left(\frac{1+x}{1-x}\right), \quad |x|<1$$ $$\coth^{-1} x = \frac12\ln\left(\frac{x+1}{x-1}\right), \quad |x|>1$$

Derivatives

$$\frac{d}{dx}\sinh^{-1}u = \frac{u'}{\sqrt{1+u^2}}, \quad \frac{d}{dx}\cosh^{-1}u = \frac{u'}{\sqrt{u^2-1}} \quad (u>1)$$ $$\frac{d}{dx}\tanh^{-1}u = \frac{u'}{1-u^2} \quad (|u|<1), \quad \frac{d}{dx}\coth^{-1}u = \frac{u'}{1-u^2} \quad (|u|>1)$$ $$\frac{d}{dx}\operatorname{sech}^{-1}u = -\frac{u'}{u\sqrt{1-u^2}} \quad (0<u<1)$$

Integrals Leading to Inverse Hyperbolic

$$\int \frac{dx}{\sqrt{x^2 + a^2}} = \sinh^{-1}\left(\frac{x}{a}\right) + C = \ln\left|x + \sqrt{x^2 + a^2}\right| + C$$ $$\int \frac{dx}{\sqrt{x^2 - a^2}} = \cosh^{-1}\left(\frac{x}{a}\right) + C = \ln\left|x + \sqrt{x^2 - a^2}\right| + C$$ $$\int \frac{dx}{a^2 - x^2} = \frac{1}{a}\tanh^{-1}\left(\frac{x}{a}\right) + C = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C \quad (|x|<a)$$


Differential Equations

Bernoulli: $\displaystyle\frac{dy}{dx} + P(x)y = Q(x)y^n \quad (n \neq 0,1)$

Steps:

  1. Divide through so coefficient of $\frac{dy}{dx}$ is 1
  2. Identify $n$, $P$, $Q$
  3. Substitute $v = y^{1-n}$ ⇒ $\frac{dv}{dx} + (1-n)Pv = (1-n)Q$
  4. Integrating factor: $\mu = e^{\int (1-n)P,dx}$
  5. Solve: $\mu v = \int \mu(1-n)Q,dx + C$
  6. Back-substitute $y^{1-n} = v$

Non-Homogeneous DE: $(a_1x + b_1y + c_1)dx + (a_2x + b_2y + c_2)dy = 0$

Linearly Independent ($a_1b_2 - a_2b_1 \neq 0$):

  1. Solve $\begin{cases} a_1h + b_1k + c_1 = 0 \ a_2h + b_2k + c_2 = 0 \end{cases}$ for $(h,k)$
  2. $x = u + h$, $y = v + k$ → get homogeneous DE in $u,v$
  3. Substitute $v = mu$, separate, integrate, back-substitute

Linearly Dependent ($a_1b_2 - a_2b_1 = 0$):

  1. Substitute $u = a_1x + b_1y$
  2. Reduce to separable equation, solve

Exact DE: $M,dx + N,dy = 0$

  • Test: $\displaystyle\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
  • Solve: $F = \int M,dx + g(y)$, then $F_y = N$ to find $g(y)$. Solution: $F(x,y) = C$

Mixing Tank

$$\frac{dA}{dt} = C_{in}R_{in} - \frac{A(t)}{V(t)}R_{out}$$ $$V(t) = V_0 + (R_{in} - R_{out})t$$


Maclaurin Series

Formula

$$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

Standard Series

Function Series (up to $x^5$)
$e^x$ $1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120}$
$\sin x$ $x - \frac{x^3}{6} + \frac{x^5}{120}$
$\cos x$ $1 - \frac{x^2}{2} + \frac{x^4}{24}$
$\tan x$ $x + \frac{x^3}{3} + \frac{2x^5}{15}$
$\arctan x$ $x - \frac{x^3}{3} + \frac{x^5}{5}$
$\ln(1+x)$ $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5}$
$\frac{1}{1-x}$ $1 + x + x^2 + x^3 + \cdots$
$\sinh x$ $x + \frac{x^3}{6} + \frac{x^5}{120} + \cdots$
$\cosh x$ $1 + \frac{x^2}{2} + \frac{x^4}{24} + \cdots$
$\sec x$ $1 + \frac{x^2}{2} + \frac{5x^4}{24} + \cdots$

Operations

  • Substitution: Replace $x$ with argument (e.g., $e^{-x^2}$)
  • Multiplication: Multiply two series term-by-term
  • Integration: Integrate term-by-term
  • Differentiation: Differentiate term-by-term

Derivatives of $\tan x$ at $x=0$ (for Maclaurin)

$$f(0)=0,; f'(0)=1,; f''(0)=0,; f'''(0)=2,; f^{(4)}(0)=0,; f^{(5)}(0)=16$$


Quick Reference — All Derivatives

$f(x)$ $f'(x)$
$\sin^{-1} x$ $\frac{1}{\sqrt{1-x^2}}$
$\cos^{-1} x$ $\frac{-1}{\sqrt{1-x^2}}$
$\tan^{-1} x$ $\frac{1}{1+x^2}$
$\sinh x$ $\cosh x$
$\cosh x$ $\sinh x$
$\tanh x$ $\text{sech}^2 x$
$\sinh^{-1} x$ $\frac{1}{\sqrt{1+x^2}}$
$\cosh^{-1} x$ $\frac{1}{\sqrt{x^2-1}}$
$\tanh^{-1} x$ $\frac{1}{1-x^2}$

FAC1004 only. Based on lectures (En Hisham Safuan), UAS 22-23/23-24/24-25, and Exam Leaks 2025-2026.