Question

put the following equation in standard complex number form (a+bi)

Answer 1

\[(\sqrt{3}/2+1/2i)^100\]

Answer 2

the the 100th power

Answer 3

do you know de movire theroerm?

Answer 4

yes i do the french guy

Answer 5

\[(\cos \theta +i*\sin \theta)^{n}=\cos(n \theta)+isin(n \theta)\]

Answer 6

yes i know about it

Answer 7

so what is your r and theta in this problem?

Answer 8

\[z=\frac{ \sqrt{3} }{ 2 } + \frac{ 1 }{ 2 }i\]

you can work out the complex number in the form of z=r(cosx+isinx)
then you know that

\[z^{n}=r^{n}(\cos(nx)+isin(nx))\]

Answer 9

okay

Answer 10

your are is sqrt(a^2+b^2) and theta is tan^-1(b/a)

Answer 11

yes i have that

Answer 12

now just do r^100 and 100*theta

Answer 13

so i square the square root of 3 over 2 and add it to one half squared

Answer 14

yes

Answer 15

then sqrt it

Answer 16

sqrt of 1 is 1 so nvrmnd

Answer 17

okay i got the square root of 1

Answer 18

which is 1 lol

Answer 19

now 1^100 is 1 , so your knew r is 1

Answer 20

new*

Answer 21

what about theta?

Answer 22

okay now theta is the arc tangent of b over a

Answer 23

yes, which is

Answer 24

the 2 cancels out leaving me with that square root of 3

Answer 25

which is 60 did i do that right

Answer 26

1.05 rad

Answer 27

1.05*100=105 rad

Answer 28

oh i was in degree mode on my calculator

Answer 29

so we multiply it by 100 i thought we take it to the 100th power

Answer 30

but 2pi rad is 0, so you have to throw 32pi away, which leaves 4.47rad

Answer 31

you multiple r and add theta to multiplay, so power r and multiple theta to power

Answer 32

ohh okay so how can i but the answer in (a+bi) form

Answer 33

well, you are in polar corrd right now, how do you coordinate transform from polar to ractangular?

Answer 34

do i have to do cosine or the square a^2 +b^2 again

Answer 35

i cant remember how :(

Answer 36

no

Answer 37

x=rcos(theta), y=rsin(theta)

Answer 38

x is your a and y is your b

Answer 39

okay so the cosine of the square root of 3 over 2

Answer 40

i got .99

Answer 41

and for the sine of the other one i got .008 i think i am doing something wrong

Answer 42

oh okay so what to do with those numbers

Answer 43

rcostheta is a, and rsin theta is b

Answer 44

do you have answr to check?

Answer 45

so -0.24 is a and -.97 is b so -0.24+-.97i

Answer 46

I hope that is right, but not 100% positive, I am too rust on complex

Answer 47

okay well thanks lol

Answer 48

do you know how to do this one : \[5i^7\]

Answer 49

5*i^7 or (5i)^7?

Answer 50

you dont need de movire for this

Answer 51

cuz i^4=1 so i^7=i^3=i^2*i=-i

Answer 52

ohhh so its i^3

Answer 53

which is i

Answer 54

but i^3 is-i

Answer 55

no, -i

Answer 56

oh oops -i

Answer 57

so -5i I guess

Answer 58

lol are you sure

Answer 59

i have an example one maybe that will help you help me better

Answer 60

sure

Answer 61

for the problem 7i^64 you divide 68 by 4 and then you get 7i^4 and that equals 7 and 7i^68 equals 7 this is a problem from the book

Answer 62

sorry the problem is 7i^68 not 64

Answer 63

says the 64+4=68 so thats how the answer is 7 so how can i solve 5i^7

Answer 64

yes, because i^(4n) is 1 where n is any postive integer

Answer 65

i^7=I^4*i^3=-i

Answer 66

soo ans is -5i

Answer 67

lol yea i see it now lol thanks

Answer 68

np