A polar coordinate system uses measurements of r, (distance from the origin) and \theta (angle from the positive horizontal axis called a polar axis). We use these measurements \left(r,\theta \right) instead of the cartesian measurements of \left(x,y \right). Incidently the cartesian coordinates are also referred to as rectangular coordinates (because the coordinate plane is a rectangular plane).

A coordinate plane with concentric circles

A polar coordinate system therefore looks different to our rectangular coordinate system. This is what it looks like.

We can convert our rectangular form of complex numbers a+bi into polar coordinates. To do so we need the distance from the origin, which is the modulus and the angle from the positive horizontal axis (this is called the argument).

Recognizing that the horizontal distance x, can be calculated using the angle \theta, and similarly for the height y, gives us the values of:

A coordinate plane with a diagonal line segment labled r = absolute value of z from the origin to a point P in the first quadrant with coordinates x,y. A horizontal line segment from the origin to the point directly under P on the x axis labeled x=r cosine theta. A vertical segment from the point on the x axis to point P labeled y = r sine theta. All of these segments combined make a right triangle. The angle at the origin is labeled theta.

x=r \cos \theta

y=r \sin \theta

So we can rewrite the complex number {z=x+iy} as z= r \cos \theta + ir \sin \theta.

We can take r out as a factor so we get z=r\left(\cos \theta + i \sin \theta \right).

When we know the rectangular coordinates of a point then we know the lengtha of the legs of the right triangle shown in the diagram. This allows us to find the modulus, r using the Pythagorean theorem.

We can find the argument, or the measure of the angle \theta using the fact that \tan \theta = \dfrac{y}{x}= \dfrac{r \sin \theta}{r \cos \theta}= \dfrac{ \sin \theta}{ \cos \theta}.

Examples

Example 1

For each of the following graphs, find the coordinates of the missing point.

A coordinate plane with the right side of the horizontal axis labeled 0, the top vertical axis labeled pi over 2. The left horizontal axis labeled pi and the bottom vertical axis labeled 3 pi over 2. There is a diagonal segment in the first quadrant drawn from the origin to the point r, theta making an angle theta with the horizontal axis marked with a double angle marking. There is a diagonal segment in the fourth quadrant to an unknown point. It makes an angle with the horizontal axis marked with a double angle marking.

Worked Solution

Create a strategy

The coordinates given, and the coordinates we need to find, are polar coordinates. We can compare the position of the coordinate we are given with the coordinate we need to find.

The new point sits on a ray the same length as the original ray making an angle the same size as the original angle but in the negative direction.

Apply the idea

\left(r, -\theta \right)

Reflect and check

Notice the original point has been reflected across the horizontal axis. This resulted in the sign of the angle changing from positive to negative.

Worked Solution

Create a strategy

The new point sits on a ray that is the same length as the original ray. The ray makes an angle with the negative horizontal axis that is the same size as the original angle. But remember, we must measure the angle starting from 0.

If we start at 0 on the horizontal axis and move to \pi we will pass the new ray by a measure of \theta so we must subtract \theta from \pi to find the measure of the new angle.

Apply the idea

\left(r, \pi - \theta \right)

Reflect and check

Notice the original point was reflected across the vertical axis. This resulted in a new angle with a measure of \pi minus the original angle. This is because a point to the left side of the horizontal axis represents an angle of \pi.

Worked Solution

Create a strategy

If we start at 0 on the horizontal axis and move to \pi we will fall short of the new ray by a measure of \theta so we must add \theta to \pi to find the measure of the new angle.

Apply the idea

\left(r, \pi + \theta \right)

Reflect and check

Notice the new point represents a rotation of the original point by 180 \degree or \pi radians. This is why we add \pi to the original angle measure.

Example 2

Consider the graph where the complex number P is plotted:

-4

-3

-2

-1

-4

-3

-2

-1

imaginary

Write P in rectangular form.

Worked Solution

Create a strategy

The horizontal axis is the x-axis and the vertical axis is the imaginary axis.

Apply the idea

P=-2+4i

Write P in polar form.

Worked Solution

Create a strategy

We need to write the coordinates of the point in the form r\left(\cos \theta + i \sin \theta \right).

Visualize the right triangle formed by point P.

-4

-3

-2

-1

-4

-3

-2

-1

imaginary

We need to find r using the Pythagorean theorem and \theta using \tan^{-1}.

Apply the idea

First we will find r.

Find the length of each leg of the right triangle and use those values in the Pythagorean theorem.

-4

-3

-2

-1

-4

-3

-2

-1

imaginary

\displaystyle r^2	\displaystyle =	\displaystyle a^2+b^2	Pythagorean theorem
\displaystyle r^2	\displaystyle =	\displaystyle \left(-2\right)^2+4^2	Substitute the lengths of the legs
\displaystyle r^2	\displaystyle =	\displaystyle 4+16	Evaluate the squares
\displaystyle r^2	\displaystyle =	\displaystyle 20	Add
\displaystyle r	\displaystyle =	\displaystyle \sqrt{20}	Take square root of both sides
\displaystyle r	\displaystyle =	\displaystyle 2\sqrt{5}	Simplify

Next we will find \theta. Let's call the reference angle between the horizontal axis and the ray \alpha.

\displaystyle \tan \alpha	\displaystyle =	\displaystyle \dfrac{r \sin \alpha}{r \cos \alpha}
\displaystyle \tan \alpha	\displaystyle =	\displaystyle \dfrac{4}{2}	Substitute
\displaystyle \tan \alpha	\displaystyle =	\displaystyle 2	Simplify
\displaystyle \alpha	\displaystyle =	\displaystyle \tan^{-1}\left(2\right)	Apply inverse tangent
\displaystyle \alpha	\displaystyle \approx	\displaystyle 1.107148718	Evaluate using a calculator
\displaystyle \theta	\displaystyle =	\displaystyle \pi - \alpha	Find \theta
\displaystyle \theta	\displaystyle \approx	\displaystyle \pi -1.107148718	Substitute
\displaystyle \theta	\displaystyle \approx	\displaystyle 2.034443936	Subtract

Writing in polar form we get 2 \sqrt{5}\left[\cos \left(2.0344\right) + i \sin \left(2.0344\right) \right].

Example 3

What are the rectangular coordinates of the point P if its polar coordinates are: \left(2,\dfrac{7\pi}{4}\right)?

Worked Solution

Create a strategy

A point with polar coordinates \left(r, \theta \right) has rectangular coordinates \left(x, y \right) where:

x=r \cos \theta \\ y=r \sin \theta

For the point P, \text{ } r=2 and \theta=\dfrac{7\pi}{4}.

Apply the idea

Find x:

\displaystyle x	\displaystyle =	\displaystyle r \cos \theta
\displaystyle x	\displaystyle =	\displaystyle 2 \cos \left(\dfrac{7\pi}{4}\right)	Substitute r and \theta
\displaystyle x	\displaystyle =	\displaystyle 2\cdot \dfrac{\sqrt{2}}{2}	Evaluate \cos
\displaystyle x	\displaystyle =	\displaystyle \sqrt{2}	Simplify

Find y:

\displaystyle y	\displaystyle =	\displaystyle r \sin \theta
\displaystyle y	\displaystyle =	\displaystyle 2 \sin \left(\dfrac{7\pi}{4}\right)	Substitute r and \theta
\displaystyle y	\displaystyle =	\displaystyle 2\cdot \dfrac{-\sqrt{2}}{2}	Evaluate \sin
\displaystyle y	\displaystyle =	\displaystyle -\sqrt{2}	Simplify

Writing as coordinates: \left(\sqrt{2},-\sqrt{2}\right).

Example 4

Write the complex number 5-5i in polar form.

Worked Solution

Create a strategy

The modulus r for a complex number x+iy is \sqrt{x^2+y^2}. For this number x=5 and y=-5.

The argument \theta satisfies the equation \tan \theta=\dfrac{y}{x}.

Apply the idea

Let's start by finding r.

\displaystyle r	\displaystyle =	\displaystyle \sqrt{x^2+y^2}
	\displaystyle =	\displaystyle \sqrt{5^2+\left(-5\right)^2}	Substitute x and y
	\displaystyle =	\displaystyle \sqrt{25+25}	Evaluate the squares
	\displaystyle =	\displaystyle \sqrt{50}	Add
	\displaystyle =	\displaystyle 5\sqrt{2}	Simplify

Next we will find \theta.

\displaystyle \tan \theta	\displaystyle =	\displaystyle \dfrac{y}{x}
\displaystyle \tan \theta	\displaystyle =	\displaystyle \dfrac{-5}{5}	Substitute x and y
\displaystyle \tan \theta	\displaystyle =	\displaystyle -1	Simplify
\displaystyle \theta	\displaystyle =	\displaystyle \tan^{-1} \left(-1\right)	Apply \tan^-1
\displaystyle \theta	\displaystyle =	\displaystyle -\dfrac{\pi}{4}	Evaluate

Substituting r=5\sqrt{2} and \theta=-\dfrac{\pi}{4} into polar form we get:

5\sqrt{2}\left[\cos \left(-\dfrac{\pi}{4}\right) + i\sin \left(-\dfrac{\pi}{4}\right)\right]

Idea summary

Polar coordinates use the variables \left(r,\theta\right) where r is the radius and \theta is the angle with the positive horizontal axis.

In the polar coordinate system:

x=r \cos \theta

y=r \sin \theta

A complex number z=x+iy can be written as z=r\left(\cos \theta +i \sin \theta \right).

The fact that \tan \theta=\dfrac{\sin \theta}{\cos \theta} can be used to solve for \theta.

Outcomes

3.13.A

Determine the location of a point in the plane using both rectangular and polar coordinates.

3.13 Trigonometry and polar coordinates

Introduction

Ideas

Polar coordinates

Examples

Example 1

Example 2

Example 3

Example 4

Outcomes

3.13.A

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