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Mathematics 26 Online
OpenStudy (anonymous):

How would you go about doing this question without a calculator. I know I have to know my unit circle for this. but yea how would one go about solving this problem without a calculator?

OpenStudy (anonymous):

OpenStudy (anonymous):

So for the first question I got into polar form using oilers formula \[e ^{\frac{ 4\pi i }{ 3 }}\] Its kind of hard to see that but its says e^(4pi(i)/3)

OpenStudy (anonymous):

\[(e ^{\frac{ 4\pi \iota }{ 3}})^{111}\]

OpenStudy (anonymous):

how would I get the exact value for this in the form a + bi

OpenStudy (anonymous):

@freckles

OpenStudy (amistre64):

put it into trig form

OpenStudy (amistre64):

at least thats my first thought

OpenStudy (amistre64):

bah, im not real sure what the question is asking for i spose ...

OpenStudy (anonymous):

its asking to raise it to the power of 111 and simply if it to put into a + bi form

OpenStudy (amistre64):

(r,t)^n = (r^n, nt) right?

OpenStudy (amistre64):

r,t being the ordered pair for (r,t) -> r(cos(t) + i sin(t))

OpenStudy (anonymous):

yes

OpenStudy (amistre64):

it appears r = 1, since its the unit circle, what is t?

OpenStudy (anonymous):

4pi/3 I believe

OpenStudy (amistre64):

something about 60 yes, and the other is something about 45

OpenStudy (amistre64):

111 (4pi/3) ... reduces to what in periodic terms?

OpenStudy (anonymous):

148pi?

OpenStudy (amistre64):

can we simplify that to between 0 and 2pi?

OpenStudy (anonymous):

thats where im getting stuck, how would you do that?

OpenStudy (amistre64):

2k = 148, when k=74 so 74 revolutions gets us to 148 right?

OpenStudy (anonymous):

yes

OpenStudy (amistre64):

then 148 pi is just a large version of 2pi

OpenStudy (amistre64):

which is just a slightly larger version of 0

OpenStudy (amistre64):

cos(0) + i sin(0)

OpenStudy (anonymous):

1+0i

OpenStudy (amistre64):

444pi/3 = 148pi = k*2pi

OpenStudy (amistre64):

yes

OpenStudy (anonymous):

what I dont understand is that how did you know 148pi is equivalent to 2pi?

OpenStudy (amistre64):

becuase a circle repeats itself every 2pi ... so 2pi = 4pi = 6pi = ... how many times does 2pi go into 148pi? how many times does the circle turn around on itself?

OpenStudy (anonymous):

74 times

OpenStudy (amistre64):

then 74 revolutions just gets us back to the start of the circle |dw:1445712174118:dw|

OpenStudy (anonymous):

ok so say the angle was 195pi how would you get the equivalent angle between 0 and 2pi?

OpenStudy (amistre64):

divide it by 2pi, whats the remainder?

OpenStudy (anonymous):

1/2

OpenStudy (amistre64):

then 1/2 of 2pi is pi

OpenStudy (amistre64):

if its not a full rotation of 2pi, then its a partial rotation ...

OpenStudy (amistre64):

after 97 rotation, we have 1/2 a rotation

OpenStudy (anonymous):

so an angle of 873 pi would also be equivalent to pi?

OpenStudy (anonymous):

cuz 873/2= 436.5. and remainder is 0.5

OpenStudy (amistre64):

spose we have 23.3247 rotations of 2pi ... then we have 23 + 0.3247 rotations ...

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