Polar Coordinates

Angle $θ$ from the positive x-axis and magnitude $r = | z |$
To convert between polar and standard coordinates, use $x = r \cos θ, y = r \sin θ$
With this, then multiplication is simply $r_{1} r_{2} (\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2}))$ , or $r_{1} r_{2} c i s (θ_{1} + θ_{2})$
- Basically, the angles are added up, and then scaled by the product of the magnitudes
Shorthand: $c i s θ = \cos θ + i \sin θ$
- $i = c i s (\frac{π}{2})$
- $c i s (θ_{1}) \cdot c i s (θ_{2}) = c i s (θ_{1} + θ_{2})$
- $\frac{c i s (θ_{1})}{c i s (θ_{2})} = c i s (θ_{1} - θ_{2})$
- $c i s (0) = 1$
- [[#De Moivre's Theorem]]
- Notice that these are properties shared by the exponential function
- So we can define $e^{i θ} = c i s θ$
  - We can prove this with Taylor Series
  - Technically you can use any base if you change $θ$
  - Simply set $θ = π$ and get very cool equation Top 1 equation
Interesting to note: $$
\begin{align}
(cis\theta)^{2} \
\begin{split}
& cis\theta cis\theta \
& =(\cos\theta+i\sin\theta)^{2} \
& =\cos ^{2}\theta-\sin ^{2}\theta+i(2\sin\theta \cos\theta)
\end{split} &\quad
\begin{split}
& cis(2\theta) \
& =\cos(2\theta)+i\sin(2\theta)
\end{split}
\end

f o r m s t h e t r i g d o u b l e a n g l e i d e n t i t i e s - S h a p e : c a r d o i d $ r = 1 + \sin θ $ ‘ ‘ ‘ d e s m o s - g r a p h r = \sin θ + 1 ‘ ‘ ‘ - T o d r a w : - T o f i n d s l o p e s o f t a n g e n t l i n e s [[F 0 - G l o s s a r y / C o n c r e t e C o n c e p t s / D e r i v a t i v e ∥ D e r i v a t i v e]], c o n v e r t t o $ \frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} $ - H o r i z o n t a l t a n g e n t s h a v e $ \frac{d y}{d θ} = 0, \frac{d x}{d θ} \neq 0 $ - V e r t i c a l t a n g e n t s a r e o p p o s i t e - I f t h e o t h e r o n e i s z e r o, y o u c a n s e e i f i t^{'} s a c t u a l l y a h o r i z o n t a l / v e r t i c a l t a n g e n t b y t a k i n g t h e l i m i t a n d a p p l y i n g [[F 0 - G l o s s a r y / C o n c r e t e C o n c e p t s / L^{'} H ô p i t a l^{'} s R u l e ∥ L^{'} H ô p i t a l^{'} s R u l e]] - T o t a k e t h e [[F 0 - G l o s s a r y / C o n c r e t e C o n c e p t s / I n t e g r a l ∥ I n t e g r a l]], f i n d t h e a r e a o f t h e c i r c l e s e c t o r w i t h r a d i u s $ r $ a n d s e c t o r a n g l e $ d θ $ - Y o u e n d u p w i t h $ d A = \frac{1}{2} r^{2} d θ = \frac{1}{2} f (x)^{2} d θ $ - T h e i n t e g r a l i s t h e n t h e s u m : $ A = \frac{1}{2} \int_{α}^{β} f (x)^{2} d x $ - F o r [[F 0 - G l o s s a r y / C o n c r e t e C o n c e p t s / A r c L e n g t h o f C u r v e ∥ A r c L e n g t h o f C u r v e]], s i m p l y d o i t a s n o r m a l, b u t u s e $ d S = \sqrt{{(\frac{d y}{d θ})}^{2} + {(\frac{d x}{d θ})}^{2}} d θ $ - N o t i c e t h a t $ x = f (θ) \cos θ ⟹ \frac{d x}{d θ} = f^{'} (θ) \cos θ - f (θ) \sin θ $ a n d $ y = f (θ) \sin θ ⟹ \frac{d y}{d θ} = f^{'} (θ) \sin θ + f (θ) \cos θ $ - T h e n

\begin{align}
& \left( \frac{ \mathrm{d}y }{ \mathrm{d}\theta } \right)^{2}+\left( \frac{ \mathrm{d}x }{ \mathrm{d}\theta } \right)^{2} \
& =f'(\theta)^{2}\cos ^{2}\theta\cancel{ -2f'(\theta)f(\theta)\cos\theta \sin\theta }+f(\theta)^{2}\sin ^{2}\theta+f'(\theta)^{2}\sin ^{2}\theta+\cancel{ 2f'(\theta)f(\theta)\cos\theta \sin\theta }+f(\theta)^{2}\cos ^{2}\theta \
& =f'(\theta)^{2}+f(\theta)^{2} \
dS & =\sqrt{ f'(\theta)^{2}+f(\theta)^{2} } d\theta
\end{align}$$

De Moivre's Theorem

$z^{n} = r^{n} c i s (n θ)$ for any integer $n$
Prove using induction and multiplicative property