Lesson Objectives

- Learn about the complex plane (Argand diagram)
- Learn how to express the sum of complex numbers graphically
- Learn how to find the trigonometric (polar) form of a complex number
- Learn how to convert between the rectangular and polar forms

## How to Work with Complex Numbers on the Argand Diagram

Previously in our course, we learned about the complex plane, which is also known as the Argand diagram: We can see from our complex plane above, we have modified our Cartesian Coordinate System by calling the x-axis the real axis and the y-axis the imaginary axis. Each complex number a + b$i$ can be graphed by starting at the origin and moving a units horizontally, followed by moving b units vertically. Let's look at an example.

Example #1: Plot the given complex number. $$-7 + 6i$$ Additionally, we can also think about graphing a complex number using a position vector. Each complex number a + b$i$ determines a unique position vector with an initial point at the origin 0 + 0$i$ (0,0) and a terminal point a + b$i$ (a,b). Let's look at an example.

Example #2: Graph the given complex number as a position vector. $$8 + 5i$$

Example #3: Find the sum of the complex numbers graphically. $$(6 + 5i) + (-9 + 3i)$$ To find our sum: $$(6 + 5i) + (-9 + 3i)$$ Add the real parts, then add the imaginary parts: $$(6 + (-9)) + (5 + 3)i=-3 + 8i$$ $$(6 + 5i) + (-9 + 3i)=-3 + 8i$$ Graphically, we can show the resultant using the parallelogram method.

Example #4: Write each in rectangular form. $$2(\text{cos}\hspace{.1em}150° + i \hspace{.1em}\text{sin}\hspace{.1em}150°)$$ Recall that our polar form is given as: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ Recall that our rectangular form is given as: $$x + yi$$ $$x=r \cdot \text{cos}\hspace{.1em}θ$$ $$y=r \cdot \text{sin}\hspace{.1em}θ$$ We know that r, the number outside of the parentheses is 2.

Let's find cos 150°, sin 150°: $$\text{cos}\hspace{.1em}150°=-\frac{\sqrt{3}}{2}$$ $$\text{sin}\hspace{.1em}150°=\frac{1}{2}$$ $$r=2$$ Putting everything together: $$2\left(-\frac{\sqrt{3}}{2}+ \frac{1}{2}i\right)$$ Distribute the 2 to each term: $$-\sqrt{3}+ i$$ Example #5: Write each in rectangular form. $$6(\text{cos}\hspace{.1em}135° + i \hspace{.1em}\text{sin}\hspace{.1em}135°)$$ Recall that our polar form is given as: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ Recall that our rectangular form is given as: $$x + yi$$ $$x=r \cdot \text{cos}\hspace{.1em}θ$$ $$y=r \cdot \text{sin}\hspace{.1em}θ$$ We know that r, the number outside of the parentheses is 6.

Let's find cos 135°, sin 135°: $$\text{cos}\hspace{.1em}135°=-\frac{\sqrt{2}}{2}$$ $$\text{sin}\hspace{.1em}135°=\frac{\sqrt{2}}{2}$$ $$r=6$$ Putting everything together: $$6\left(-\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i\right)$$ Distribute the 6 to each term: $$-3\sqrt{2}+ 3i\sqrt{2}$$

Example #6: Write each in polar form. $$2 - 2i\sqrt{3}$$ Step 1: Let's think about the quadrant of our complex number.

Our real part is 2, while the imaginary part is -2$\sqrt{3}$. This means we would move from the origin 2 units to the right, followed by 2$\sqrt{3}$ units down. This would put our complex number in quadrant IV.

Step 2: Find the value of r, this is the absolute value of our complex number. $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{2^2 + (-2\sqrt{3})^{2}}$$ $$r=\sqrt{4 + (4 \cdot 3)}$$ $$r=\sqrt{4 + 12}$$ $$r=\sqrt{16}$$ $$r=4$$ Step 3: Find θ. $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ $$\text{tan}\hspace{.1em}θ=-\frac{2\sqrt{3}}{2}$$ $$\text{tan}\hspace{.1em}θ=-\sqrt{3}$$ Since we know we are in quadrant IV, let's find our reference angle: $$\text{tan}^{-1}(\sqrt{3})=60°$$ In quadrant IV, we would subtract our reference angle from 360°: $$θ=360° - 60°=300°$$ Now, we are ready to write our polar form: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ $$r=4, θ=300°$$ $$4 (\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300°)$$ $$4\hspace{.1em}\text{cis}\hspace{.1em}300°$$ Example #7: Write each in polar form. $$-\frac{\sqrt{31}}{2}+ \frac{\sqrt{93}}{2}i$$ Step 1: Let's think about the quadrant of our complex number.

Our real part is -$\frac{\sqrt{31}}{2}$, while the imaginary part is $\frac{\sqrt{93}}{2}$. This means we would move from the origin $\frac{\sqrt{31}}{2}$ units to the left, followed by $\frac{\sqrt{93}}{2}$ units up. This would put our complex number in quadrant II.

Step 2: Find the value of r, this is the absolute value of our complex number. $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{\left(-\frac{\sqrt{31}}{2}\right)^2 + \left(\frac{\sqrt{93}}{2}\right)^2}$$ $$r=\sqrt{\frac{31}{4}+ \frac{93}{4}}$$ $$r=\sqrt{\frac{124}{4}}$$ $$r=\sqrt{31}$$ Step 3: Find θ. $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ $$\text{tan}\hspace{.1em}θ=-\Large{\frac{\frac{\sqrt{93}}{2}}{\frac{\sqrt{31}}{2}}}$$ $$\text{tan}\hspace{.1em}θ=-\frac{\sqrt{93}}{\sqrt{31}}$$ $$\text{tan}\hspace{.1em}θ=-\sqrt{3}$$ Since we know we are in quadrant II, let's find our reference angle: $$\text{tan}^{-1}(\sqrt{3})=60°$$ In quadrant II, we would subtract our reference angle from 180°: $$θ=180° - 60°=120°$$ Now, we are ready to write our polar form: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ $$r=\sqrt{31}, θ=120°$$ $$\sqrt{31}(\text{cos}\hspace{.1em}120° + i \hspace{.1em}\text{sin}\hspace{.1em}120°)$$ $$\sqrt{31}\hspace{.1em}\text{cis}\hspace{.1em}120°$$ Example #8: Write each in polar form. $$-2\sqrt{3}- 2i$$ Step 1: Let's think about the quadrant of our complex number.

Our real part is -2$\sqrt{3}$, while the imaginary part is -2. This means we would move from the origin $2\sqrt{3}$ units to the left, followed by 2 units down. This would put our complex number in quadrant III.

Step 2: Find the value of r, this is the absolute value of our complex number. $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{(-2\sqrt{3})^{2}+ (-2)^2}$$ $$r=\sqrt{(4 \cdot 3) + 4}$$ $$r=\sqrt{12 + 4}$$ $$r=\sqrt{16}$$ $$r=4$$ Step 3: Find θ. $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ $$\text{tan}\hspace{.1em}θ=\frac{-2}{-2\sqrt{3}}$$ $$\text{tan}\hspace{.1em}θ=\frac{1}{\sqrt{3}}$$ $$\text{tan}\hspace{.1em}θ=\frac{\sqrt{3}}{3}$$ Since we know we are in quadrant III, let's find our reference angle: $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=30°$$ In quadrant III, we would add 180° to our reference angle: $$θ=180° + 30°=210°$$ Now, we are ready to write our polar form: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ $$r=4, θ=210°$$ $$4 (\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$ $$4 \hspace{.1em}\text{cis}\hspace{.1em}210°$$

Example #1: Plot the given complex number. $$-7 + 6i$$ Additionally, we can also think about graphing a complex number using a position vector. Each complex number a + b$i$ determines a unique position vector with an initial point at the origin 0 + 0$i$ (0,0) and a terminal point a + b$i$ (a,b). Let's look at an example.

Example #2: Graph the given complex number as a position vector. $$8 + 5i$$

### Expressing the Sum of Two Complex Numbers Graphically

We previously learned how to find the sum or difference of two complex numbers. Graphically, the sum of two complex numbers is represented by the vector that is the resultant of the vectors corresponding to the two complex numbers. Let's look at an example.Example #3: Find the sum of the complex numbers graphically. $$(6 + 5i) + (-9 + 3i)$$ To find our sum: $$(6 + 5i) + (-9 + 3i)$$ Add the real parts, then add the imaginary parts: $$(6 + (-9)) + (5 + 3)i=-3 + 8i$$ $$(6 + 5i) + (-9 + 3i)=-3 + 8i$$ Graphically, we can show the resultant using the parallelogram method.

## How to Convert Complex Numbers to Polar Form

### Trigonometric (Polar) Form of a Complex Number

When our complex number is expressed in the form of a + b$i$, it is known as the rectangular form (standard form) of a complex number. Let's consider a different form known as the trigonometric form or polar form of a complex number. Instead of a as our real part, we will use x, and instead of b as our imaginary part, we will use y. Our image above shows the complex number x + y$i$ as a position vector. Our vector has a direction angle θ and a magnitude r. Let's consider the relationships between x, y, r, and θ: $$x=r \hspace{.1em}\text{cos}\hspace{.1em}θ$$ $$y=r \hspace{.1em}\text{sin}\hspace{.1em}θ$$ $$r=\sqrt{x^2 + y^2}$$ $$\text{tan}\hspace{.1em}θ=\frac{y}{x}, x ≠ 0$$ If we plug in r cos θ for x and r sin θ for y into x + y$i$, we obtain the following: $$x + yi=r \hspace{.1em}\text{cos}\hspace{.1em}θ + (r \hspace{.1em}\text{sin}\hspace{.1em}θ)i$$ Now, let's factor out the r: $$r \hspace{.1em}\text{cos}\hspace{.1em}θ + (r \hspace{.1em}\text{sin}\hspace{.1em}θ)i=r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ Our given expression is known as the trigonometric form or the polar form of a complex number: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ In some cases, this will be abbreviated using: $$r \hspace{.1em}\text{cis}\hspace{.1em}θ$$ The number r is the absolute value of our complex number x + y$i$. $$r=\sqrt{x^2 + y^2}$$ θ is known as the argument of our complex number x + y$i$. For our purposes, we will choose θ to be in the interval: [0°, 360°), however, any coterminal angle with θ could also serve as our argument.### Converting from Trigonometric (Polar) Form to Rectangular Form

It is very easy to convert from polar form to rectangular form. Let's look at a few examples.Example #4: Write each in rectangular form. $$2(\text{cos}\hspace{.1em}150° + i \hspace{.1em}\text{sin}\hspace{.1em}150°)$$ Recall that our polar form is given as: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ Recall that our rectangular form is given as: $$x + yi$$ $$x=r \cdot \text{cos}\hspace{.1em}θ$$ $$y=r \cdot \text{sin}\hspace{.1em}θ$$ We know that r, the number outside of the parentheses is 2.

Let's find cos 150°, sin 150°: $$\text{cos}\hspace{.1em}150°=-\frac{\sqrt{3}}{2}$$ $$\text{sin}\hspace{.1em}150°=\frac{1}{2}$$ $$r=2$$ Putting everything together: $$2\left(-\frac{\sqrt{3}}{2}+ \frac{1}{2}i\right)$$ Distribute the 2 to each term: $$-\sqrt{3}+ i$$ Example #5: Write each in rectangular form. $$6(\text{cos}\hspace{.1em}135° + i \hspace{.1em}\text{sin}\hspace{.1em}135°)$$ Recall that our polar form is given as: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ Recall that our rectangular form is given as: $$x + yi$$ $$x=r \cdot \text{cos}\hspace{.1em}θ$$ $$y=r \cdot \text{sin}\hspace{.1em}θ$$ We know that r, the number outside of the parentheses is 6.

Let's find cos 135°, sin 135°: $$\text{cos}\hspace{.1em}135°=-\frac{\sqrt{2}}{2}$$ $$\text{sin}\hspace{.1em}135°=\frac{\sqrt{2}}{2}$$ $$r=6$$ Putting everything together: $$6\left(-\frac{\sqrt{2}}{2}+ \frac{\sqrt{2}}{2}i\right)$$ Distribute the 6 to each term: $$-3\sqrt{2}+ 3i\sqrt{2}$$

### Converting from Rectangular Form to Trigonometric (Polar) Form

- Determine the quadrant of the complex number x + y$i$
- It may help to create a simple sketch and graph the complex number

- Find the value of r (the absolute value of the complex number)
- Find θ, using tan θ = y/x, x ≠ 0
- Choose the correct quadrant based on step 1

Example #6: Write each in polar form. $$2 - 2i\sqrt{3}$$ Step 1: Let's think about the quadrant of our complex number.

Our real part is 2, while the imaginary part is -2$\sqrt{3}$. This means we would move from the origin 2 units to the right, followed by 2$\sqrt{3}$ units down. This would put our complex number in quadrant IV.

Step 2: Find the value of r, this is the absolute value of our complex number. $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{2^2 + (-2\sqrt{3})^{2}}$$ $$r=\sqrt{4 + (4 \cdot 3)}$$ $$r=\sqrt{4 + 12}$$ $$r=\sqrt{16}$$ $$r=4$$ Step 3: Find θ. $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ $$\text{tan}\hspace{.1em}θ=-\frac{2\sqrt{3}}{2}$$ $$\text{tan}\hspace{.1em}θ=-\sqrt{3}$$ Since we know we are in quadrant IV, let's find our reference angle: $$\text{tan}^{-1}(\sqrt{3})=60°$$ In quadrant IV, we would subtract our reference angle from 360°: $$θ=360° - 60°=300°$$ Now, we are ready to write our polar form: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ $$r=4, θ=300°$$ $$4 (\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300°)$$ $$4\hspace{.1em}\text{cis}\hspace{.1em}300°$$ Example #7: Write each in polar form. $$-\frac{\sqrt{31}}{2}+ \frac{\sqrt{93}}{2}i$$ Step 1: Let's think about the quadrant of our complex number.

Our real part is -$\frac{\sqrt{31}}{2}$, while the imaginary part is $\frac{\sqrt{93}}{2}$. This means we would move from the origin $\frac{\sqrt{31}}{2}$ units to the left, followed by $\frac{\sqrt{93}}{2}$ units up. This would put our complex number in quadrant II.

Step 2: Find the value of r, this is the absolute value of our complex number. $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{\left(-\frac{\sqrt{31}}{2}\right)^2 + \left(\frac{\sqrt{93}}{2}\right)^2}$$ $$r=\sqrt{\frac{31}{4}+ \frac{93}{4}}$$ $$r=\sqrt{\frac{124}{4}}$$ $$r=\sqrt{31}$$ Step 3: Find θ. $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ $$\text{tan}\hspace{.1em}θ=-\Large{\frac{\frac{\sqrt{93}}{2}}{\frac{\sqrt{31}}{2}}}$$ $$\text{tan}\hspace{.1em}θ=-\frac{\sqrt{93}}{\sqrt{31}}$$ $$\text{tan}\hspace{.1em}θ=-\sqrt{3}$$ Since we know we are in quadrant II, let's find our reference angle: $$\text{tan}^{-1}(\sqrt{3})=60°$$ In quadrant II, we would subtract our reference angle from 180°: $$θ=180° - 60°=120°$$ Now, we are ready to write our polar form: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ $$r=\sqrt{31}, θ=120°$$ $$\sqrt{31}(\text{cos}\hspace{.1em}120° + i \hspace{.1em}\text{sin}\hspace{.1em}120°)$$ $$\sqrt{31}\hspace{.1em}\text{cis}\hspace{.1em}120°$$ Example #8: Write each in polar form. $$-2\sqrt{3}- 2i$$ Step 1: Let's think about the quadrant of our complex number.

Our real part is -2$\sqrt{3}$, while the imaginary part is -2. This means we would move from the origin $2\sqrt{3}$ units to the left, followed by 2 units down. This would put our complex number in quadrant III.

Step 2: Find the value of r, this is the absolute value of our complex number. $$r=\sqrt{x^2 + y^2}$$ $$r=\sqrt{(-2\sqrt{3})^{2}+ (-2)^2}$$ $$r=\sqrt{(4 \cdot 3) + 4}$$ $$r=\sqrt{12 + 4}$$ $$r=\sqrt{16}$$ $$r=4$$ Step 3: Find θ. $$\text{tan}\hspace{.1em}θ=\frac{y}{x}$$ $$\text{tan}\hspace{.1em}θ=\frac{-2}{-2\sqrt{3}}$$ $$\text{tan}\hspace{.1em}θ=\frac{1}{\sqrt{3}}$$ $$\text{tan}\hspace{.1em}θ=\frac{\sqrt{3}}{3}$$ Since we know we are in quadrant III, let's find our reference angle: $$\text{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right)=30°$$ In quadrant III, we would add 180° to our reference angle: $$θ=180° + 30°=210°$$ Now, we are ready to write our polar form: $$r (\text{cos}\hspace{.1em}θ + i \hspace{.1em}\text{sin}\hspace{.1em}θ)$$ $$r=4, θ=210°$$ $$4 (\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$ $$4 \hspace{.1em}\text{cis}\hspace{.1em}210°$$

#### Skills Check:

Example #1

Write each in rectangular form. $$6(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°)$$

Please choose the best answer.

A

$$-3\sqrt{2}- 3i\sqrt{2}$$

B

$$\sqrt{17}- 4i\sqrt{17}$$

C

$$4 + i\sqrt{5}$$

D

$$\sqrt{2}+ i\sqrt{2}$$

E

$$-\frac{3\sqrt{2}}{2}- \frac{3\sqrt{2}}{2}i$$

Example #2

Write each in polar form. $$-\frac{\sqrt{105}}{2}- \frac{\sqrt{35}}{2}i$$

Please choose the best answer.

A

$$2(\text{cos}\hspace{.1em}90° + i \hspace{.1em}\text{sin}\hspace{.1em}90°)$$

B

$$2(\text{cos}\hspace{.1em}0° + i \hspace{.1em}\text{sin}\hspace{.1em}0°)$$

C

$$\sqrt{35}(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$

D

$$\sqrt{14}(\text{cos}\hspace{.1em}240° + i \hspace{.1em}\text{sin}\hspace{.1em}240°)$$

E

$$5(\text{cos}\hspace{.1em}140° + i \hspace{.1em}\text{sin}\hspace{.1em}140°)$$

Example #3

Write each in polar form. $$\sqrt{10}+ i \hspace{.1em}\sqrt{10}$$

Please choose the best answer.

A

$$\sqrt{14}(\text{cos}\hspace{.1em}225° + i \hspace{.1em}\text{sin}\hspace{.1em}225°)$$

B

$$4(\text{cos}\hspace{.1em}300° + i \hspace{.1em}\text{sin}\hspace{.1em}300°)$$

C

$$4(\text{cos}\hspace{.1em}90° + i \hspace{.1em}\text{sin}\hspace{.1em}90°)$$

D

$$2\sqrt{5}(\text{cos}\hspace{.1em}45° + i \hspace{.1em}\text{sin}\hspace{.1em}45°)$$

E

$$\sqrt{2}(\text{cos}\hspace{.1em}210° + i \hspace{.1em}\text{sin}\hspace{.1em}210°)$$

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