Question
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Convert the rectangular form of the complex number 2−2i
into polar form. Show all work and label the modulus and argument.
Answers
GPT-4o mini
To convert the complex number \( z = 2 - 2i \) from rectangular form to polar form, we need to find its modulus and argument.
### Step 1: Calculate the modulus
The modulus \( r \) of a complex number \( z = a + bi \) is given by the formula:
\[
r = \sqrt{a^2 + b^2}
\]
For our complex number \( z = 2 - 2i \), we have:
- \( a = 2 \)
- \( b = -2 \)
Now we can substitute the values into the formula:
\[
r = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
### Step 2: Calculate the argument
The argument \( \theta \) of a complex number \( z = a + bi \) can be found using:
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
For \( z = 2 - 2i \):
- \( a = 2 \)
- \( b = -2 \)
Calculating the argument:
\[
\theta = \tan^{-1}\left(\frac{-2}{2}\right) = \tan^{-1}(-1)
\]
The value of \( \tan^{-1}(-1) \) is \( -\frac{\pi}{4} \) radians. However, since the point \( (2, -2) \) is in the fourth quadrant, we can express the angle as:
\[
\theta = 2\pi - \frac{\pi}{4} = \frac{8\pi}{4} - \frac{\pi}{4} = \frac{7\pi}{4}
\]
### Step 3: Write the polar form
The polar form of a complex number is given by:
\[
z = r (\cos \theta + i \sin \theta)
\]
So, substituting the modulus \( r = 2\sqrt{2} \) and the argument \( \theta = \frac{7\pi}{4} \):
\[
z = 2\sqrt{2} \left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right)
\]
This can also be expressed in exponential form using Euler's formula:
\[
z = r e^{i\theta} = 2\sqrt{2} e^{i\frac{7\pi}{4}}
\]
### Final Answer
The polar form of the complex number \( 2 - 2i \) is:
\[
z = 2\sqrt{2} \left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right)
\]
or
\[
z = 2\sqrt{2} e^{i\frac{7\pi}{4}}
\]
where the modulus is \( 2\sqrt{2} \) and the argument is \( \frac{7\pi}{4} \).
### Step 1: Calculate the modulus
The modulus \( r \) of a complex number \( z = a + bi \) is given by the formula:
\[
r = \sqrt{a^2 + b^2}
\]
For our complex number \( z = 2 - 2i \), we have:
- \( a = 2 \)
- \( b = -2 \)
Now we can substitute the values into the formula:
\[
r = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
### Step 2: Calculate the argument
The argument \( \theta \) of a complex number \( z = a + bi \) can be found using:
\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]
For \( z = 2 - 2i \):
- \( a = 2 \)
- \( b = -2 \)
Calculating the argument:
\[
\theta = \tan^{-1}\left(\frac{-2}{2}\right) = \tan^{-1}(-1)
\]
The value of \( \tan^{-1}(-1) \) is \( -\frac{\pi}{4} \) radians. However, since the point \( (2, -2) \) is in the fourth quadrant, we can express the angle as:
\[
\theta = 2\pi - \frac{\pi}{4} = \frac{8\pi}{4} - \frac{\pi}{4} = \frac{7\pi}{4}
\]
### Step 3: Write the polar form
The polar form of a complex number is given by:
\[
z = r (\cos \theta + i \sin \theta)
\]
So, substituting the modulus \( r = 2\sqrt{2} \) and the argument \( \theta = \frac{7\pi}{4} \):
\[
z = 2\sqrt{2} \left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right)
\]
This can also be expressed in exponential form using Euler's formula:
\[
z = r e^{i\theta} = 2\sqrt{2} e^{i\frac{7\pi}{4}}
\]
### Final Answer
The polar form of the complex number \( 2 - 2i \) is:
\[
z = 2\sqrt{2} \left(\cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right)\right)
\]
or
\[
z = 2\sqrt{2} e^{i\frac{7\pi}{4}}
\]
where the modulus is \( 2\sqrt{2} \) and the argument is \( \frac{7\pi}{4} \).