1. The limaçon represented by r=3-2sinθ can be classified as a limaçon of Pascal. It is a closed curve with a single loop, and the loop is not centered at the origin.
2. The quantity of petals in r=3cos9θ is 9, and the length of each petal is 1.
3. First, we simplify z1/z2:
z1/z2 = (6(cos45°+isin45°))/(2(cos90°+isin90°))
= 3(cos45°+isin45°)(cos(-90°)+isin(-90°))
= 3(cos135°+isin135°)
Then, we can write z1/z2 in polar form:
z1/z2 = 3(cos135°+isin135°) = 3∠135°
4. We can first write 216(cos315°+isin315°) in polar form:
216(cos315°+isin315°) = 216∠315°
Then, we can find the cube roots by taking the cube root of the magnitude and dividing the angle by 3:
cube root of 216∠315° = 6∠105°
So the three cube roots are:
6(cos105°+isin105°)
6(cos(105°+120°)+isin(105°+120°)) = 6(cos225°+isin225°)
6(cos(105°+240°)+isin(105°+240°)) = 6(cos345°+isin345°)
1.Sketch and classify the limaꞔon represented by r = 3 − 2 sin θ
2.What is the quantity and the length of the petals in r = 3 cos 9θ?
3.Given z1 = 6(cos 45° + i sin 45°) and z2 = 2(cos 90° + i sin 90°), find z1/z2 when 0 ≤ θ ≤ 360°
4.Find the three cube roots of 216(cos 315 + i sin 315)
1 answer
1 answer