15) One possible equation for a rose curve with the given specifications is:
x = 5 cos(8θ)
This equation produces a curve that repeats itself 8 times for every full rotation (2π) around the origin, creating a symmetrical pattern of petals with a length of 5 units.
16) One possible equation for a lemniscate with a loop length of 7 and symmetry about the pole (x-axis) is:
r^2 = 7^2 cos(2θ)
This equation produces a curve that resembles a figure eight, with the center at the origin and loops that touch the x-axis. The loop length is determined by the value of the parameter a in the equation, which is 7 in this case.
19 Part A) To find the product of two complex numbers, we can use the formula:
z₁ * z₂ = |z₁| |z₂| cis(θ₁ + θ₂)
where |z| is the magnitude (or absolute value) of z, and θ is the argument (or angle) of z measured counterclockwise from the positive real axis, in radians.
First, let's write z₁ and z₂ in polar form:
z₁ = 1/2 cis(π/3) = 1/2 (cos π/3 + i sin π/3) = 1/2 + i(√3/2)
z₂ = 3 (cos π/6 + i sin π/6) = 3/2 + i(3/2)
Then, we can calculate:
|z₁| = sqrt((1/2)^2 + (√3/2)^2) = √(1/4 + 3/4) = 1
|z₂| = sqrt((3/2)^2 + (3/2)^2) = 3/2 √2
θ₁ = π/3 and θ₂ = π/6
Therefore:
z₁ * z₂ = |z₁| |z₂| cis(θ₁ + θ₂) = 1 * (3/2 √2) cis(π/3 + π/6)
= (3/2) √2 cis(2π/6) = (3/2) √2 cisπ/3
19 Part B) To express the complex number in rectangular (or Cartesian) form, we use the formula:
z = x + yi
where x is the real part (or the horizontal coordinate) and y is the imaginary part (or the vertical coordinate) of the complex number.
From the polar form of z₁ * z₂, we know that the magnitude of the product is (3/2) √2, and the argument is π/3. Therefore:
x = |z₁ * z₂| cos(π/3) = (3/2) √2 cos(π/3) = (3/4) √6
y = |z₁ * z₂| sin(π/3) = (3/2) √2 sin(π/3) = (3/4) √6
Hence, the rectangular form of z₁ * z₂ is:
z₁ * z₂ = (3/4) √6 + i(3/4) √6
15) Write a rose curve equation that lies on the horizontal axis, has a petal length of 5, and has 8 petals.
16) Write a lemniscate equation that is symmetric about the pole and has a loop length of 7.
19 Part A) Find the product for the equations below. z₁ * z₂ (Use exact values, no decimals. Show your work)
z₁= 1/2(cis pi/3) and z₂=3 (cos pi/6 + i sin pi/6)
19 Part B) Place your answer from Part A in rectangular form. (Use exact values and show your work)
3 answers
15) r = cos(nθ) has 2n petals when n is even. So, the curve we want is
r = 5cos(4θ)
16) the arc length of a lemniscate is an elliptic integral, so this will not work.
19) π/3 + π/6 = π/2, so
1/2(cis π/3) * 3 (cis π/6) = 3/2 cis(π/2) = 3/2 i
r = 5cos(4θ)
16) the arc length of a lemniscate is an elliptic integral, so this will not work.
19) π/3 + π/6 = π/2, so
1/2(cis π/3) * 3 (cis π/6) = 3/2 cis(π/2) = 3/2 i
19 Part B) To convert 3/2i to rectangular form, we can write:
3/2i = 0 + (3/2)i
Therefore, the rectangular form of z₁ * z₂ is:
z₁ * z₂ = 0 + (3/2)i
3/2i = 0 + (3/2)i
Therefore, the rectangular form of z₁ * z₂ is:
z₁ * z₂ = 0 + (3/2)i
1 answer