5.6.1 Complex Number System 🔧
| Euler's Formula |
$e^{\pm i\cdot\theta}=\cos(\theta)\pm i\cdot\sin(\theta),$
$\forall\theta\isin\R$
|
| Form Equivalences |
|
| Natural Logarithms of Negatives |
$\ln(-x)=i\cdot\pi+\ln(x),$
$\forall x\gt 0\isin\R$ |
| Natural Logarithms of Imaginaries |
$\ln(i)=\frac{\pi}{2}\cdot i$ |
Proof of Euler's Formula
Let $e^{\pm i\cdot\theta}=\cos(\theta)\pm i\cdot\sin(\theta)$ so that
$$f(\theta)=\frac{\cos(\theta)\pm i\cdot\sin(\theta)}{e^{\pm i\cdot\theta}}$$
Negate the exponent
$$f(\theta)=e^{\mp i\cdot\theta}\cdot\big(\cos(\theta)\pm i\cdot\sin(\theta)\big)$$
Derive using the
derivative product rule
Proof of Negative Natural Logarithms
Proof of Imaginary Natural Logarithms
5.6.2 Analytic Function for $\sin(\theta)$
$$\sin(\theta)=\frac{e^{i\cdot\theta}-e^{-i\cdot\theta}}{2\cdot i}$$
Proof
Subtract the negative exponent version of
Euler's formula from itself
$$e^{i\cdot\theta}-e^{-i\cdot\theta}=\cos(\theta)+i\cdot\sin(\theta)-\cos(\theta)+i\cdot\sin(\theta)$$
Simplify
$$e^{i\cdot\theta}+e^{-i\cdot\theta}=2\cdot i\cdot\sin(\theta)$$
Isolate sine
5.6.3 Analytic Function for $\cos(\theta)$
$$\cos(\theta)=\frac{e^{i\cdot\theta}+e^{-i\cdot\theta}}{2}$$
Proof
Add the negative exponent version of
Euler's formula to itself
$$e^{i\cdot\theta}+e^{-i\cdot\theta}=\cos(\theta)+i\cdot\sin(\theta)+\cos(\theta)-i\cdot\sin(\theta)$$
Simplify
$$e^{i\cdot\theta}+e^{-i\cdot\theta}=2\cdot\cos(\theta)$$
Isolate cosine
5.6.4 Analytic Function for $\sec(\theta)$
$$\sec(\theta)=\frac{2}{e^{i\cdot\theta}+e^{-i\cdot\theta}}$$
Proof
Equate secant to its
cofunction
$$\sec(\theta)=\frac{1}{\cos(\theta)}$$
Substitute cosine for its
analytic function
$$\sec(\theta)=1\bigg/\bigg(\frac{e^{i\cdot\theta}+e^{-i\cdot\theta}}{2}\bigg)$$
Apply the reciprocal
rule of division
5.6.5 Analytic Function for $\csc(\theta)$
$$\csc(\theta)=\frac{2\cdot i}{e^{i\cdot\theta}-e^{-i\cdot\theta}}$$
Proof
Equate cosecant to its
cofunction
$$\csc(\theta)=\frac{1}{\sin(\theta)}$$
Substitute sine for its
analytic function
$$\csc(\theta)=1\bigg/\bigg(\frac{e^{i\cdot\theta}-e^{-i\cdot\theta}}{2\cdot i}\bigg)$$
Apply the reciprocal
rule of division
5.6.6 Analytic Function for $\tan(\theta)$
$$\tan(\theta)=\frac{e^{i\cdot\theta}-e^{-i\cdot\theta}}{i\cdot\big(e^{i\cdot\theta}+e^{-i\cdot\theta}\big)}$$
Proof
Substitute the
cofunctions of tangent for their analytic functions
$$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\Bigg(\frac{e^{i\cdot\theta}-e^{-i\cdot\theta}}{2\cdot i}\bigg/\frac{e^{i\cdot\theta}+e^{-i\cdot\theta}}{2}\Bigg)$$
Simplify
5.6.7 Analytic Function for $\cot(\theta)$
$$\cot(\theta)=\frac{i\cdot\big(e^{i\cdot\theta}+e^{-i\cdot\theta}\big)}{e^{i\cdot\theta}-e^{-i\cdot\theta}}$$
Proof
Substitute the
cofunctions of cotangent for their analytic functions
$$\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}=\Bigg(\frac{e^{i\cdot\theta}+e^{-i\cdot\theta}}{2}\bigg/\frac{e^{i\cdot\theta}-e^{-i\cdot\theta}}{2\cdot i}\Bigg)$$
Simplify
5.6.8 Analytic Function for $\sin^{-1}(x)$ 🔧
$$\sin^{-1}(x)=-i\cdot\ln\big(i\cdot x+\sqrt{1-x^2}\big)$$
Proof
5.6.9 Analytic Function for $\cos^{-1}(x)$ 🔧
$$\cos^{-1}(x)=-i\cdot\ln\big(x+i\cdot\sqrt{1-x^2}\big)$$
Proof
5.6.A Analytic Function for $\csc^{-1}(x)$ 🔧
$$\csc^{-1}(x)=-i\cdot\ln\big(i\cdot x^{-1}+\sqrt{1-x^{-2}}\big)$$
Proof
5.6.B Analytic Function for $\sec^{-1}(x)$ 🔧
$$\sec^{-1}(x)=-i\cdot\ln\big(x^{-1}+i\cdot\sqrt{1-x^{-2}}\big)$$
Proof
5.6.C Analytic Function for $\tan^{-1}(x)$ 🔧
$$\tan^{-1}(x)=\frac{i}{2}\cdot\ln\bigg(\frac{1-i\cdot x}{1+i\cdot x}\bigg)$$
Proof
5.6.D Analytic Function for $\cot^{-1}(x)$ 🔧
$$\cot^{-1}(x)=\frac{i}{2}\cdot\ln\bigg(\frac{i\cdot x+1}{i\cdot x-1}\bigg)$$
Proof
