Tag: Polar Coordinates

Convert complex number in polar form to the form a + bi

by RoRi

February 27, 2016

Convert each of the following complex numbers given in polar form to the form $a+bi$ .

$e^{\frac{\pi i}{2}}$ .
$2e^{-\frac{\pi i}{2}}$ .
$3e^{\pi i}$ .
$-e^{-\pi i}$ .
$i + e^{2 \pi i}$ .
$e^{\frac{\pi i}{4}}$ .
$e^{\frac{\pi i}{4}} - e^{-\frac{\pi i}{4}}$ .
$\displaystyle{ \frac{1 - e^{\frac{\pi i}{2}}}{1 + e^{\frac{ \pi i}{2}}}}$ .

Using the definition of the complex exponential ( $e^{i \theta} = \cos \theta + i \sin \theta$ ) we have
$e^{ \frac{\pi i}{2}} = \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} = 0 + i = i.$
Again, using the definition of the complex exponential we have,
$2e^{-\frac{\pi i}{2}} = 2 \left(\cos \frac{-\pi}{2} + i \sin \frac{-\pi}{2} \right)= -2i.$
We have
$3e^{\pi i} = 3 \left( \cos \pi + i \sin \pi \right) = -3.$
We have
$-e^{-\pi i} = - \left( \cos (-\pi) + i \sin (-\pi) \right) = 1.$
We have
$i + e^{2 \pi i} = i + \cos (2 \pi) + i \sin (2 \pi) = i + 1.$
We have
$e^{ \frac{\pi i}{4}} = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} = \frac{1 +i}{\sqrt{2}}.$
We have
$e^{\frac{\pi i}{4}} - e^{-\frac{\pi i}{4}} = \frac{1+i}{\sqrt{2}} - \frac{1-i}{\sqrt{2}} = \sqrt{2} i.$
We have
$\frac{1 - e^{\frac{\pi i}{2}}}{1 + e^{\frac{\pi i}{2}}} = \frac{1-i}{1+i} = \frac{-2i}{2} = -i.$

Sketch a “limacon” and compute its area from 0 to 2 π

by RoRi

August 9, 2015

Define a limacon by:

$f(\theta) = 2+\cos \theta, \qquad 0 \leq \theta \leq 2 \pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{2 \pi} (4 + 4 \cos \theta + \cos^2 \theta) \, d \theta &= 2 \int_0^{2 \pi} d \theta + 2 \int_0^{2 \pi} \cos \theta \, d \theta + \frac{1}{2} \int_0^{2 \pi} \cos^2 \theta \, d \theta \\ &= 4 \pi + 0 + \frac{\pi}{2} \\ &= \frac{9 \pi}{2}. \end{align*}$

Sketch a “cardioid” and compute its area from 0 to 2 π

by RoRi

August 9, 2015

Define a cardioid by:

$f(\theta) = 1+\cos \theta, \qquad 0 \leq \theta \leq 2 \pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{2 \pi} (1 + 2 \cos \theta + \cos^2 \theta) \, d \theta &= \frac{1}{2} \int_0^{2 \pi} d \theta + \int_0^{2 \pi} \cos \theta \, d \theta + \frac{1}{2} \int_0^{2 \pi} \cos^2 \theta \, d\theta \\ &= \pi + 0 + \frac{\pi}{2} \\ &= \frac{3 \pi}{2} \end{align*}$

where we know $\int_0^{2 \pi} \cos^2 \theta \, d \theta = \pi$ from this exercise (Section 2.11, Exercise #7).

Sketch a “four-leaf clover” and compute its area from 0 to 2 π

by RoRi

August 9, 2015

Define a four-leaf clover by:

$f(\theta) = \sqrt{|\cos \theta|}, \qquad 0 \leq \theta \leq 2 \pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{2 \pi} | \cos (2 \theta) | \, d \theta &= \frac{1}{4} \int_0^{4 \pi} | \cos \theta| \, d \theta \\ &= \frac{1}{2} \int_0^{2 \pi} | \cos \theta | \, d \theta \\ &= 2, \end{align*}$

where for the final step we used the previous exercise (Section 2.11, Exercise #12).

Sketch a “lazy eight” and compute its area from 0 to 2π

by RoRi

August 8, 2015

Define a lazy eight by:

$f(\theta) = \sqrt{|\cos \theta|}, \qquad 0 \leq \theta \leq 2 \pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{2 \pi} | \cos \theta | \, d \theta &= \frac{1}{2} \left( \int_0^{\frac{\pi}{2}} \cos \theta \, d \theta - \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos \theta \, d \theta + \int_{\frac{3 \pi}{2}}^{2 \pi} \cos \theta \, d\theta \right) \\ &= \frac{1}{2} (1 - (-1) + 1 - (-1)) \\ &= 2. \end{align*}$

Sketch a “four-leaved rose” and compute its area from 0 to 2π

by RoRi

August 8, 2015

Define a four-leaved rose by:

$f(\theta) = |\sin (2 \theta)|, \qquad 0 \leq \theta \leq 2 \pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{2 \pi} \sin^2 (2 \theta) \, d \theta &= \frac{1}{4} \int_0^{4 \pi} \sin^2 \theta \, d \theta \\ &= \frac{1}{8} \int_0^{4 \pi} d \theta - \frac{1}{8} \int_0^{4 \pi} \cos (2 \theta) \, d \theta \\ &= \frac{\pi}{2} - \frac{1}{16} \int_0^{8 \pi} \cos \theta \, d \theta \\ &= \frac{\pi}{2}. \end{align*}$

Sketch a “rose petal” and compute its area from 0 to π/2

by RoRi

August 8, 2015

Define a rose petal by:

$f(\theta) = \sin (2 \theta), \qquad 0 \leq \theta \leq \pi/2.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{\frac{\pi}{2}} \sin^2 (2 \theta) \, d \theta &= \frac{1}{8} \int_0^{\pi} 2 \sin^2 \theta \, d \theta \\ &= \frac{1}{8} \int_0^{\pi} d \theta - \frac{1}{8} \int_0^{\pi} \cos (2 \theta) \, d \theta \\ &= \frac{\pi}{8} - \frac{1}{16} \int_0^{2 \pi} \cos \theta \, d \theta \\ &= \frac{\pi}{8}. \end{align*}$

Sketch two circles tangent to the x-axis and compute their area from 0 to 2π

by RoRi

August 8, 2015

Define two circles tangent to the $x$ -axis by:

$f(\theta) = 4 |\sin \theta|, \qquad 0 \leq \theta \leq 2\pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{2 \pi} 16 \sin^2 \theta \, d \theta &= 4 \int_0^{2 \pi} d \theta - 4 \int_0^{2 \pi} \cos (2 \theta) \, d \theta \\ &= 8 \pi. \end{align*}$

Sketch a circle tangent to the x-axis and compute its area from 0 to π

by RoRi

August 8, 2015

Define a circle tangent to the $x$ -axis by:

$f(\theta) = 4 \sin \theta, \qquad 0 \leq \theta \leq \pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{\pi} 16 \sin^2 \theta \, d \theta &= -4 \int_0^{\pi} (1 - 2 \sin^2 \theta - 1) \, d \theta \\ &= 4 \int_0^{\pi} d \theta - 4 \int_0^{\pi} \cos (2 \theta) \, d \theta \\ &= 4 \pi - 2 \int_0^{2 \pi} \cos \theta \, d \theta \\ &= 4 \pi. \end{align*}$

Sketch two circles tangent to the y-axis and compute their area from 0 to 2 π

by RoRi

August 7, 2015

Define two circles tangent to the $y$ -axis by:

$f(\theta) = 2 |\cos \theta|, \qquad 0 \leq \theta \leq 2 \pi.$

Sketch this graph in polar coordinates and compute the area of the radial set.

The sketch is as follows:

Rendered by QuickLaTeX.com

Then, we compute the area,

$\begin{align*} a(R) = \frac{1}{2} \int_0^{2 \pi} 4 \cos^2 \theta \, d \theta &= 2\int_0^{2 \pi} \left( \frac{1 + \cos (2\theta)}{2} \right) \, d \theta \\ &= \int_0^{2 \pi} d \theta + \int_0^{2 \pi} \cos (2 \theta) \, d \theta \\ &= 2 \pi + 0 = 2 \pi. \end{align*}$