To solve for [tex]\( \frac{z_1}{z_2} = a + bi \)[/tex] where [tex]\( z_1 = 2 \operatorname{cis} 120^{\circ} \)[/tex] and [tex]\( z_2 = 4 \operatorname{cis} 30^{\circ} \)[/tex], follow these steps:
1. Find the magnitude of [tex]\( \frac{z_1}{z_2} \)[/tex]:
- The magnitude of [tex]\( z_1 \)[/tex] is 2.
- The magnitude of [tex]\( z_2 \)[/tex] is 4.
- The magnitude of [tex]\( \frac{z_1}{z_2} \)[/tex] is given by:
[tex]\[
\text{Magnitude} = \frac{2}{4} = 0.5
\][/tex]
2. Find the angle of [tex]\( \frac{z_1}{z_2} \)[/tex]:
- The angle of [tex]\( z_1 \)[/tex] is [tex]\( 120^\circ \)[/tex].
- The angle of [tex]\( z_2 \)[/tex] is [tex]\( 30^\circ \)[/tex].
- The angle of [tex]\( \frac{z_1}{z_2} \)[/tex] is:
[tex]\[
\text{Angle} = 120^\circ - 30^\circ = 90^\circ
\][/tex]
3. Express [tex]\( \frac{z_1}{z_2} \)[/tex] in rectangular form [tex]\( a + bi \)[/tex]:
- Use the magnitude (0.5) and the angle (90°) to find [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
\frac{z_1}{z_2} = 0.5 \operatorname{cis} 90^\circ
\][/tex]
- Convert [tex]\( 0.5 \operatorname{cis} 90^\circ \)[/tex] to rectangular form using [tex]\( \operatorname{cis} \theta = \cos \theta + i \sin \theta \)[/tex]:
[tex]\[
a = \text{Magnitude} \times \cos(\text{Angle}) = 0.5 \times \cos(90^\circ) = 0.5 \times 0 = 0
\][/tex]
[tex]\[
b = \text{Magnitude} \times \sin(\text{Angle}) = 0.5 \times \sin(90^\circ) = 0.5 \times 1 = 0.5
\][/tex]
However, we see that the cosine part is close to zero, and thus, the number reveals an actual result close to zero in practical terms.
Therefore, the values are:
[tex]\[ a = 3.061616997868383 \times 10^{-17} \approx 0 \][/tex]
[tex]\[ b = 0.5 \][/tex]
Thus,
[tex]\[ \boxed{0} \][/tex]
and
[tex]\[ \boxed{0.5} \][/tex]
