Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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**Problem Statement:**

Find the product \( z_1z_2 \) and the quotient \( \frac{z_1}{z_2} \). Express your answers in polar form.

Given:

\[ z_1 = 10(\cos(110^\circ) + i \sin(110^\circ)) \]

\[ z_2 = 2(\cos(20^\circ) + i \sin(20^\circ)) \]

**Solution:**

\[ z_1z_2 = \_\_\_\_ \]

\[ \frac{z_1}{z_2} = \_\_\_\_ \]

**Explanation:**

In polar form, the complex number \( z = r(\cos \theta + i \sin \theta) \) can be treated using the modulus \( r \) and argument \( \theta \).

For the product \( z_1z_2 \), multiply the moduli and add the arguments:

\[ z_1z_2 = (10 \times 2)\left(\cos(110^\circ + 20^\circ) + i \sin(110^\circ + 20^\circ)\right) = 20(\cos(130^\circ) + i \sin(130^\circ)) \]

For the quotient \( \frac{z_1}{z_2} \), divide the moduli and subtract the arguments:

\[ \frac{z_1}{z_2} = \left(\frac{10}{2}\right)\left(\cos(110^\circ - 20^\circ) + i \sin(110^\circ - 20^\circ)\right) = 5(\cos(90^\circ) + i \sin(90^\circ)) \]

\( \cos(90^\circ) = 0 \) and \( \sin(90^\circ) = 1 \), so:

\[ \frac{z_1}{z_2} = 5i \]
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Transcribed Image Text:**Problem Statement:** Find the product \( z_1z_2 \) and the quotient \( \frac{z_1}{z_2} \). Express your answers in polar form. Given: \[ z_1 = 10(\cos(110^\circ) + i \sin(110^\circ)) \] \[ z_2 = 2(\cos(20^\circ) + i \sin(20^\circ)) \] **Solution:** \[ z_1z_2 = \_\_\_\_ \] \[ \frac{z_1}{z_2} = \_\_\_\_ \] **Explanation:** In polar form, the complex number \( z = r(\cos \theta + i \sin \theta) \) can be treated using the modulus \( r \) and argument \( \theta \). For the product \( z_1z_2 \), multiply the moduli and add the arguments: \[ z_1z_2 = (10 \times 2)\left(\cos(110^\circ + 20^\circ) + i \sin(110^\circ + 20^\circ)\right) = 20(\cos(130^\circ) + i \sin(130^\circ)) \] For the quotient \( \frac{z_1}{z_2} \), divide the moduli and subtract the arguments: \[ \frac{z_1}{z_2} = \left(\frac{10}{2}\right)\left(\cos(110^\circ - 20^\circ) + i \sin(110^\circ - 20^\circ)\right) = 5(\cos(90^\circ) + i \sin(90^\circ)) \] \( \cos(90^\circ) = 0 \) and \( \sin(90^\circ) = 1 \), so: \[ \frac{z_1}{z_2} = 5i \]
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