2007-06-08 13:05:20 · 5 answers · asked by boswelltrinity 1 in Science & Mathematics ➔ Mathematics
You're trying to convert z = r(cos(theta) + i sin(theta)) to z = a + bi. You do this in that a = r cos(theta) and b = r sin(theta).
Thus, the answer is 6 + 10.392i , or 6 + 12 (sqrt(3)/2) .
2007-06-08 13:16:00 · answer #1 · answered by Michael S 2 · 0⤊ 0⤋
cos Ï/3 = 1/2
sin Ï/3 = â3/2
Gives:-
12.(1/2 + i â3/2)
= 6.(1 + â3 i)
2007-06-09 03:26:08 · answer #2 · answered by Como 7 · 0⤊ 0⤋
sin(pi/3)=sqrt(3)/2
cos(pi/3)=1/2
Thus, your number becomes:
12[cos(pe/3)+isin(pi/3)]=
=12[1/2+isqrt(3)/2]=
=6+6sqrt(3)i
a=6, b=6sqrt(3)
2007-06-08 20:18:58 · answer #3 · answered by brigitte 2 · 0⤊ 0⤋
sin Ï/3=â3/2
cos Ï/3=0.5
12(cos Ï/3+isin Ï/3)=6+6iâ3
2007-06-08 20:53:11 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
cos(pi/3)+isin(pi/3)=exp(ipi/3)
==1/2+isqrt(3/2)
so 12*[1/2+isqrt(3)/2]=
6+6*sqrt(3)
2007-06-08 20:44:18 · answer #5 · answered by jon d 3 · 0⤊ 0⤋