/sci/ - Science & Math

>>10336546
[math]
\begin{alignat*}{1}
e^x &= \sum_{k=0}^{\infty} \frac{x^k}{k!} \\
&= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \\
\therefore e^{ix} &= \sum_{k=0}^{\infty} \frac{\left(ix\right)^k}{k!} \\
&= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \\
\cos{x} &= \sum_{k=0}^{\infty} \frac{\left(-1\right)^{k}x^{2k}}{\left(2k\right)!} \\
&= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots \\
\sin{x} &= \sum_{k=0}^{\infty} \frac{\left(-1\right)^{k}x^{2k+1}}{\left(2k+1\right)!} \\
&= x - \frac{x^3}{3!} + \dots \\
\therefore i\sin{x} &= i\sum_{k=0}^{\infty} \frac{\left(-1\right)^{k}x^{2k+1}}{\left(2k+1\right)!} \\
&= ix - \frac{ix^3}{3!} + \dots \\
\cos{x} + i\sin{x} &= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \dots \\
&= e^{ix} \\
e^{ix} &= \cos{x} + i\sin{x} \quad \square \\
&\text{It follows that any nonzero complex number } z \text{ can be represented in standard form as } a + bi \text{ or in polar form as } re^{i\theta} \text{.} \\
z = a + bi &= re^{i\theta} \\
&= r\left(\cos{\theta} + i\sin{\theta}\right) \\
a &= r\cos{\theta} \\
b &= r\sin{\theta} \\
\frac{b}{a} &= \frac{\sin{\theta}}{\cos{\theta}} \\
&= \tan{\theta} \\
\arctan{\left(\frac{b}{a}\right)} &= \theta \\
a^2 + b^2 &= r^2\cos^2{\theta} + r^2\sin^2{\theta} \\
&= r^2\left(\cos^2{\theta} + \sin^2{\theta}\right) \\
&= r^2 \\
\sqrt{a^2 + b^2} &= r \\
z = a + bi &= \sqrt{a^2 + b^2}e^{i\arctan{\left(\frac{b}{a}\right)}} \quad \square \\
&\text{It follows that any nonzero complex number } z \text{, being the product of a real constant } r \text{ and some power of } e \text{, has some natural logarithm } \ln{z} \text{.} \\
\ln{z} = \ln{\left(a + bi\right)} &= \ln{\left(\sqrt{a^2 + b^2}e^{i\arctan{\left(\frac{b}{a}\right)}}\right)} \\
&= \ln{\sqrt{a^2 + b^2}} + i\arctan{\left(\frac{b}{a}\right)} \quad \square \\
\end{alignat*}
[/math]