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1. find the product of
[4(cos30° + isin30°)] * [6(cos120° + isin120°)]

2. Find the the three cube roots of 1.
I don't understand, isn't the answer supposed to be one?

3. Write all solutions of the equation
x^5 + 243 = 0

answers:
1. -12√3 + 12i

2. cos0 + isin0
cos 2pi/3 + isin 2pi/3
cos 4pi/3 + isin 4pi/3

3. 3(cos pi/5 + isin pi/5)
3(cos 3pi/5 + isin 3pi/5)
3(cos pi + isin pi)
3(cos 7pi/5 + isin 7pi/5)
3(cos 9pi/5 + isin 9pi/5)

2007-12-08 23:44:52 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

1)=24 e^i30 *e ^120i =24 e ^150i = 24 (cos 150+i sin 150)=
-12sqrt3+12
Work in trigonometric form
1=1> +2kpi so 1^1/3 = 1<2k pi/3 ( k= 0,1,2)
1<0 = cos 0 +i sin 0
1<2pi/3 = cos 2pi/3 +i sin 2pi/3
and cos 4pi/3+i sin 4pi/3
3Also in trigonometric form
x^5 = -243 = 243 3(cos pi/5+i sinpi/5)
3(cos3pi/5+i sin 3pi/5) You can write the missing for k= 2,3,4

2007-12-09 00:14:32 · answer #1 · answered by santmann2002 7 · 0 0

1 and 1 and 1

-1 and -1 and 1

try multiplying those together (for a better understanding)


243= 3^5

x^5 = -243

is -3


(keep it simple)

2007-12-09 02:00:19 · answer #2 · answered by Anonymous · 0 1

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