SOLUTION: Let z = 4(cos 3π/2 + i sin 3π/2) and w = 3(cos π/6 + i sin π/6). Write the rectangular form of z/w.

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Question 1032628: Let z = 4(cos 3π/2 + i sin 3π/2) and w = 3(cos π/6 + i sin π/6). Write the rectangular form of z/w.
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
Let z = 4(cos 3π/2 + i sin 3π/2)
-----
z = 4(0 + i*-1)
z = 4(-i)
z = 0-4i
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w = 3(cos π/6 + i sin π/6).
w = 3(sqrt(3)/2 + i(1/2))
w = (3sqrt(3) + 3i)/2
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Write the rectangular form of z/w.
z/w = -4i/(3sqrt(3)+3i)/2 = -8i/(3sqrt(3) + 3i)
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= -8i(3sqrt(3)- 3i)/(27 +9)
====
= (-24 -24sqrt(3)i)/36
----
= (-2/3)(1 + sqrt(3)i)
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cheers,
Stan H.
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Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Let z = 4(cos 3π/2 + i sin 3π/2) and w = 3(cos π/6 + i sin π/6). Write the rectangular form of z/w.
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z = 4cis(3pi/2), w = 3cis(pi/6)
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z/w = (4/3)cis(3pi/2 - pi/6) = (4/3)cis(4pi/3)
z/w = (4/3)*cos(4pi/3) + (4i/3)*sin(4pi/3)
z/w = (4/3)*(-1/2) + (4i/3)*(-sqrt(3)/2)
z/w = -2/3 - i*2sqrt(3)/3
= (2/3)*(-1 - sqrt(3)i)
--> same answer as Stanbon
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