To find the product [tex]\( z_1 z_2 \)[/tex] where
[tex]\[
z_1 = 3\left(\cos 130^\circ + i \sin 130^\circ \right)
\][/tex]
and
[tex]\[
z_2 = 7\left(\cos 90^\circ + i \sin 90^\circ \right),
\][/tex]
you can use the properties of multiplication for complex numbers in polar form.
1. Find the magnitudes:
The magnitude of the product [tex]\( z_1 z_2 \)[/tex] is the product of the magnitudes of [tex]\( z_1 \)[/tex] and [tex]\( z_2 \)[/tex]:
[tex]\[
\text{Magnitude of } z_1 = 3
\][/tex]
[tex]\[
\text{Magnitude of } z_2 = 7
\][/tex]
Therefore,
[tex]\[
\text{Magnitude of } z_1 z_2 = 3 \times 7 = 21
\][/tex]
2. Find the angles:
The angle of the product [tex]\( z_1 z_2 \)[/tex] is the sum of the angles of [tex]\( z_1 \)[/tex] and [tex]\( z_2 \)[/tex]:
[tex]\[
\text{Angle of } z_1 = 130^\circ
\][/tex]
[tex]\[
\text{Angle of } z_2 = 90^\circ
\][/tex]
Therefore,
[tex]\[
\text{Angle of } z_1 z_2 = 130^\circ + 90^\circ = 220^\circ
\][/tex]
3. Combine the magnitudes and angles:
Therefore,
[tex]\[
z_1 z_2 = 21 \left( \cos 220^\circ + i \sin 220^\circ \right)
\][/tex]
Given the multiple choices:
- [tex]\( z_1 z_2 = 21 (\cos 220^\circ + i \sin 220^\circ) \)[/tex]
- [tex]\( z_1 z_2 = 21 (\cos 40^\circ + i \sin 40^\circ) \)[/tex]
- [tex]\( z_1 z_2 = \frac{3}{7} (\cos 40^\circ + i \sin 40^\circ) \)[/tex]
- [tex]\( z_1 z_2 = \frac{3}{7} (\cos 220^\circ + i \sin 220^\circ) \)[/tex]
We see that the correct representation of the product [tex]\( z_1 z_2 \)[/tex] must match our result. Thus, the correct answer is:
[tex]\[
z_1 z_2 = 21 \left( \cos 220^\circ + i \sin 220^\circ \right)
\][/tex]
