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.
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)
5 answers
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
= 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.
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!
0 answers