-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)
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)
1 answer
1 answer