Question

find the trigonometric form of -12-12(square root)3i

a. 24(cos2pi/3 + isin 2pi/3)
b. 24(cos4pi/3 + isin 4pi/3)
c. 12(cos 4pi/3 + isin 4pi/3)
d. 12(cos12pi/3 + isin 2pi/3)
e. 12(squareroot) 2 (cos4pi/3 + isin 4pi/3)

Answers

Reiny
-12-12√3 i

magnitude = √( 144 + 432)
= √576 = 24

tan Ø = -12√3/-12) = √3
Ø = 60° or π/3
but from my sketch I see that we are in quadrant III
so Ø = π/3 + π = 4π/3

-12-12√3 i
= 24(cos 4π/3 + i sin 4π/3) or 24 cis 4π/3

looks like b)

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