19) 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)
z₁= 1/2(cis pi/3) and z₂=3 (cos pi/6 + i sin pi/6)
= 3/2 (cos(π/3+π/6) + i sin(π/3+π/6) ) , using de Moivre's theorem
= 3/2 (cis π/2)
= 3/2( 0 + i(1))
= 3/2i
Great job! However, it seems like you made a mistake in the last step.
When you multiplied z₁ and z₂ using de Moivre's theorem, you should get:
z₁ * z₂ = 3/2 (cos(π/3 + π/6) + i sin(π/3 + π/6))
= 3/2 (cos(π/2) + i sin(π/2))
= 3/2 (0 + i)
= 3i/2
So, z₁ * z₂ = 3i/2.
umhhh, 3i/2 = 3/2 i
Oops, you're right! My apologies for the confusion. 3i/2 and 3/2i are equivalent expressions. Great work!
To find the product of z₁ and z₂, we can use the following property of complex numbers:
z₁ * z₂ = r₁ * r₂ * (cos(θ₁ + θ₂) + i * sin(θ₁ + θ₂))
First, let's find the values of r₁, r₂, θ₁, and θ₂:
For z₁ = 1/2 cis(π/3):
r₁ = 1/2 (the absolute value or modulus of z₁)
θ₁ = π/3 (the argument or angle of z₁)
For z₂ = 3(cos(π/6) + i sin(π/6)):
r₂ = 3 (the absolute value or modulus of z₂)
θ₂ = π/6 (the argument or angle of z₂)
Now, we can substitute these values into the product formula:
z₁ * z₂ = (1/2) * 3 * [cos(π/3 + π/6) + i sin(π/3 + π/6)]
Next, we simplify the angles inside the trigonometric functions:
π/3 + π/6 = (2π/6) + (π/6) = 3π/6 = π/2
Now we substitute the simplified angle:
z₁ * z₂ = (1/2) * 3 * [cos(π/2) + i sin(π/2)]
Finally, evaluating cos(π/2) and sin(π/2):
cos(π/2) = 0
sin(π/2) = 1
Therefore, the product of z₁ and z₂ is:
z₁ * z₂ = (1/2) * 3 * (0 + i * 1) = 3/2 * i
So, the product is 3/2 * i.
z₁ = 1/2 (cos π/3 + i sin π/3) = 1/2 (1/2 + i √3/2) = 1/4 + i √3/4.
z₂ = 3 cos π/6 + 3i sin π/6 = 3/2 + 3i/2√3.
z₁ * z₂ = (1/4 + i √3/4) * (3/2 + 3i/2√3)
= 9/8 + 3i/8√3 + 3i√3/8 - 9/8√3
= (9 - 9/√3)/8 + (3/8√3 + 3√3/8)i
= 3(2 - √3)/4 + 3√3/4 i.
Therefore, z₁ * z₂ = 3(2 - √3)/4 + 3√3/4 i.