Write 
z1 and z2
 in polar form.
z1 = −12,    z2 = 

3
 + i
z1
=	
  
z2
=	
  

Check your radius and your angle.
Find the product 
z1z2
 and the quotients 
z1
z2
 and 
1
z1
.
 Express your answers in polar form.
z1z2
=	
 
z1
z2
=	
 
1
z1
=	


To express \( z_1 \) and \( z_2 \) in polar form, let's start by identifying the radius \( r \) and the angle \( \theta \) for each.

### 1. Writing \( z_1 \) and \( z_2 \) in Polar Form

#### For \( z_1 = -12 \):
- **Radius \( r_1 \)**:
  \[
  r_1 = |z_1| = |-12| = 12
  \]
- **Angle \( \theta_1 \)**:
  Since \( z_1 \) is on the negative real axis, \( \theta_1 = \pi \) radians (or 180°).

Therefore:
\[
z_1 = 12 \text{cis}(\pi) \quad \text{where } \text{cis}(\theta) = \cos(\theta) + i\sin(\theta)
\]

#### For \( z_2 = 3 + i \):
- **Radius \( r_2 \)**:
  \[
  r_2 = |z_2| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}
  \]
- **Angle \( \theta_2 \)**:
  \[
  \theta_2 = \tan^{-1}\left(\frac{1}{3}\right)
  \]
  This angle is in the first quadrant.

Thus:
\[
z_2 = \sqrt{10} \text{cis}(\tan^{-1}(1/3))
\]

### 2. Finding the Product \( z_1z_2 \)
The product in polar form can be calculated using:
\[
z_1 z_2 = r_1 r_2 \text{cis}(\theta_1 + \theta_2)
\]
- **\( r_1 r_2 \)**:
  \[
  r_1 r_2 = 12 \cdot \sqrt{10}
  \]
- **\( \theta_1 + \theta_2 \)**:
  \[
  \theta_1 + \theta_2 = \pi + \tan^{-1}(1/3)
  \]

Therefore:
\[
z_1 z_2 = 12\sqrt{10} \text{cis}\left(\pi + \tan^{-1}(1/3)\right)
\]

### 3. Finding the Quotient \( \frac{z_1}{z_2} \)
The quotient in polar form is calculated using:
\[
\frac{z_1}{z_2} = \frac{r_1}{r_2} \text{cis}(\theta_1 - \theta_2)
\]
- **\( \frac{r_1}{r_2} \)**:
  \[
  \frac{r_1}{r_2} = \frac{12}{\sqrt{10}}
  \]
- **\( \theta_1 - \theta_2 \)**:
  \[
  \theta_1 - \theta_2 = \pi - \tan^{-1}(1/3)
  \]

Thus:
\[
\frac{z_1}{z_2} = \frac{12}{\sqrt{10}} \text{cis}\left(\pi - \tan^{-1}(1/3)\right)
\]

### 4. Finding \( \frac{1}{z_1} \)
We can find the reciprocal of \( z_1 \) as:
\[
\frac{1}{z_1} = \frac{1}{r_1} \text{cis}(-\theta_1)
\]
- **\( \frac{1}{r_1} \)**:
  \[
  \frac{1}{r_1} = \frac{1}{12}
  \]
- **\( -\theta_1 \)**:
  \[
  -\theta_1 = -\pi
  \]

Therefore:
\[
\frac{1}{z_1} = \frac{1}{12} \text{cis}(-\pi)
\]

### Final Results:
\[
z_1 = 12 \text{cis}(\pi)
\]
\[
z_2 = \sqrt{10} \text{cis}(\tan^{-1}(1/3))
\]
\[
z_1 z_2 = 12\sqrt{10} \text{cis}\left(\pi + \tan^{-1}(1/3)\right)
\]
\[
\frac{z_1}{z_2} = \frac{12}{\sqrt{10}} \text{cis}\left(\pi - \tan^{-1}(1/3)\right)
\]
\[
\frac{1}{z_1} = \frac{1}{12} \text{cis}(-\pi)
\]

Would you like more details on any part of this solution or have any questions?

### Related Questions:
1. How can we convert angles from radians to degrees?
2. What is the significance of the polar form in complex number calculations?
3. How do you determine the argument \( \theta \) for complex numbers in different quadrants?
4. Why is the product of two complex numbers simply the product of their magnitudes and the sum of their angles?
5. What are some real-world applications of complex numbers in polar form?

### Tip:
When calculating the argument \( \theta \) for a complex number, always check the quadrant to avoid incorrect angle measurements.

Write z1 and z2 in polar form. z1 = −12, z2 = 3 + i. Check your radius and your angle. Find the product z1z2 and the quotients z1/z2 and 1/z1. Express your answers in polar form.

Explore how to express complex numbers z1 = -12 and z2 = 3 + i in polar form, including detailed steps to calculate the product z1z2, the quotient z1/z2, and the reciprocal 1/z1. Learn key concepts like the polar representation of complex numbers, Euler's Formula, and De Moivre's Theorem. Suitable for students in Grades 10-12.

Math Problem Statement

z2

Check your radius and your angle. Find the product z1z2 and the quotients z1 z2 and 1 z1 . Express your answers in polar form. z1z2

z1 z2

1 z1

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