Let's start with the problem where \( z_1 = 2 \operatorname{cis} 120^{\circ} \) and \( z_2 = 4 \operatorname{cis} 30^{\circ} \). We are asked to find the values of \( a \) and \( b \) in the rectangular form of the complex number \( \frac{z_1}{z_2} = a + bi \).
Given the two complex numbers, let's express them in polar form:
- \( z_1 = 2 \operatorname{cis} 120^{\circ} \)
- \( z2 = 4 \operatorname{cis} 30^{\circ} \)
We can divide these two complex numbers in polar form by dividing their magnitudes and subtracting their angles:
[tex]\[
\frac{z_1}{z_2} = \frac{2 \operatorname{cis} 120^{\circ}}{4 \operatorname{cis} 30^{\circ}}
= \frac{2}{4} \operatorname{cis} (120^{\circ} - 30^{\circ})
= \frac{1}{2} \operatorname{cis} 90^{\circ}
\][/tex]
Next, let's convert this result back to rectangular form. The angle \( 90^{\circ} \) corresponds to \( \frac{\pi}{2} \) radians.
For a complex number \( r \operatorname{cis} \theta \) in its rectangular form, we use the relations:
[tex]\[ \operatorname{cis} \theta = \cos \theta + i \sin \theta \][/tex]
Therefore:
[tex]\[ \frac{1}{2} \operatorname{cis} 90^{\circ} = \frac{1}{2} (\cos 90^{\circ} + i \sin 90^{\circ}) \][/tex]
Knowing the values of the trigonometric functions at \( 90^{\circ} \):
- \( \cos 90^{\circ} = 0 \)
- \( \sin 90^{\circ} = 1 \)
Thus:
[tex]\[ \frac{1}{2} (0 + i \cdot 1) = \frac{1}{2} i \][/tex]
From this result, we can see that the real part \( a \) is:
[tex]\[ a = 0 \][/tex]
And the imaginary part \( b \) is:
[tex]\[ b = \frac{1}{2} \][/tex]
Therefore, the values are:
[tex]\[ a = 3.061616997868383e-17 \][/tex]
[tex]\[ b = 0.5 \][/tex]
