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

forms the trig double angle identities - Shape: cardoid $r=1+\sin\theta$ ```desmos-graph r=\sin\theta+1 ``` - To draw: - To find slopes of tangent linesDerivative, convert to $\frac{ \mathrm{d}y }{ \mathrm{d}x }=\frac{\frac{ \mathrm{d}y }{ \mathrm{d}\theta }}{\frac{ \mathrm{d}x }{ \mathrm{d}\theta }}$ - Horizontal tangents have $\frac{ \mathrm{d}y }{ \mathrm{d}\theta }=0,\frac{ \mathrm{d}x }{ \mathrm{d}\theta }\neq 0$ - Vertical tangents are opposite - If the both are zero, you can see if it's actually a horizontal/vertical tangent by taking the limit and applyingL'Hôpital's Rule- To take theIntegral, find the area of the circle sector with radius $r$ and sector angle $d\theta$ - You end up with $dA=\frac{1}{2}r^{2}d\theta=\frac{1}{2}f(\theta)^{2}d\theta$ - The integral is then the sum: $A=\frac{1}{2}\int_{\alpha}^{\beta} f(x)^{2} \, dx$ - Alternatively, simply do aAntiderivative#Double Integralssub, especially if the integrand has a $x^{2}+y^{2}$ - In this case, theJacobiansimply becomes r - This actually counters the effect of lower r points being counted more - ForArc Length of Curve, simply do it as normal, but use $dS=\sqrt{ \left( \frac{ \mathrm{d}y }{ \mathrm{d}\theta } \right)^{2}+\left( \frac{ \mathrm{d}x }{ \mathrm{d}\theta } \right)^{2} }d\theta$ - Notice that $x=f(\theta)\cos \theta\implies \frac{ \mathrm{d}x }{ \mathrm{d}\theta }=f'(\theta)\cos\theta-f(\theta)\sin\theta$ and $y=f(\theta)\sin\theta\implies \frac{ \mathrm{d}y }{ \mathrm{d}\theta }=f'(\theta)\sin\theta+f(\theta)\cos\theta$ - Then

\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