Question

4 part question

Answer 1

r = sqrt(x^2+y^2)

theta = arctan(y/x)

Answer 2

sqrt((sqrt3)^2+1^2)

theta = arctan(1/(sqrt3))

Answer 3

sqrt((sqrt3)^2+1^2)=2

theta = arctan(1/(sqrt3))=pi/6

Answer 4

have to consider the sign

it's sqrt(3) - i so the y = -1

Answer 5

sqrt((sqrt3)^2+(-1)^2)=2

theta = arctan(-1/(sqrt3))=-pi/6

Answer 6

for part I

Answer 7

good

since we want theta to be positive we will add 2pi as usual to get 11pi/6

from there you will write the polar form as

r(cos(theta)+isin(theta))

plug in your r and theta value, don't do anything else

Answer 8

r(cos(11pi/6)+isin(11pi/6))

Answer 9

don't forget to plug in the r value

Answer 10

2(cos(theta)+isin(theta))

Answer 11

ok but both theta and r need to be plugged in so plug in 11pi/6 for theta

Answer 12

2(cos(11pi/6)+isin(11pi/6))

Answer 13

good, now repeat this process w/ w

Answer 14

is this still part I or II

Answer 15

we are still on part I

Answer 16

sqrt((-2)^2+(-2)^2)=2sqrt2
theta = arctan(-2/-2)=pi/4=9pi/4

Answer 17

since we have a positive theta value there's no need to add pi this time

Answer 18

so you have r = 2sqrt(2) and theta = pi/4

then simply write this in the form r(cos(theta) + isin(theta))

Answer 19

sqrt((-2)^2+(-2)^2)=2sqrt2
theta = arctan(-2/-2)=pi/4

2sqrt2(cos(pi/4)+isin(pi/4))

Answer 20

good, then you just need to use de moivres product rule to find zw

Answer 21

4sqrt2=(Cos25pi12)+(isin25pi12)

Answer 22

the 4sqrt(2) needs to be on the right side

zw = 4sqrt(2)[(Cos25pi/12)+i(sin25pi/12)]

Answer 23

okay so is that the final for part II:

zw = 4sqrt(2)[(Cos25pi/12)+i(sin25pi/12)]

Answer 24

yes

Answer 25

then for z^2 you just apply the product formula again to z

Answer 26

2(1-isqrt3)

Answer 27

the polar form

Answer 28

apply de moivres theorem to the polar form of z to get z^2

Answer 29

the polar form is sqrt3-i righT

Answer 30

that is the rectangular form

Answer 31

-I + sqrt3?

Answer 32

refer to your calculations from part I where we calculated the polar form of z

Answer 33

2(cos(11pi/6)+isin(11pi/6))
2sqrt2(cos(pi/4)+isin(pi/4))

Answer 34

good, now simply use the product formula to calculate z^2

Answer 35

4sqrt2=[cos(25pi/12)]+[isin(25pi/12)]

Answer 36

that is the value of zw

it is asking for z^2

Answer 37

oh okay

Answer 38

(2(cos(11pi/6)+isin(11pi/6)))^2

=4(sqrt3/2+-i/2)^2

Answer 39

in polar form not rectangular

Answer 40

apply the product formula to the polar form of z

Answer 41

well the radius is four and theta is -60

Answer 42

z = 2(cos(11pi/6)+isin(11pi/6))

use the product formula to calculate z^2

Answer 43

2(sqrt3/2+-i/2)

Answer 44

:/

Answer 45

that is still rectangular form

Answer 46

oh r=

Answer 47

4[cos(11pi/6)+isin(11pi/6)]

Answer 48

notice how the theta values are being added

Answer 49

4[cos(11pi/3)+isin(11pi/3)]

Answer 50

good so that's z^2

then for part IV you will take the polar form of w, apply de moivre's theorem to find w^4

Answer 51

2sqrt2(cos(pi/4)+isin(pi/4)) so this is the original one

Answer 52

yes

Answer 53

r = 2sqrt(2)
n is the number of the exponent, so n = 4
theta = pi/4

calculate r^n * [cos(n*theta) + i*sin(n*theta)]

Answer 54

now I need to 2sqrt2(11pi/3)=2sqrt2(cos(n)(11pi/3)+isin(n*11pi/3)

Answer 55

like that

Answer 56

where are you getting 11pi/3?

w = 2sqrt2(cos(pi/4)+isin(pi/4))

so what is your theta?

Answer 57

oh I did 11pi/6 +11pi/6 that's why

Answer 58

2+2i

Answer 59

if the general form of a polar root is

z = r(cos(theta) + isin(theta))

what is the theta associated with w = 2sqrt2(cos(pi/4)+isin(pi/4))?

Answer 60

pi/4

Answer 61

good, so put everything together and find r^n * [cos(n*theta) + i*sin(n*theta)]

Answer 62

[cos(n*pi/4) + i*sin(n*pi/4)]

Answer 63

don't forget about the r^n part

Answer 64

r^n * [cos(n*pi/4) + i*sin(n*pi/4)]

Answer 65

so this gives me

Answer 66

you don't need to solve it, you just need to plug in the appropriate values of r, n, and theta

Answer 67

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Vocaloid
r = 2sqrt(2)
n is the number of the exponent, so n = 4
theta = pi/4

calculate r^n * [cos(n*theta) + i*sin(n*theta)]
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 68

2sqrt(2)^4 * [cos(4*pi/4) + i*sin(4*pi/4)]

Answer 69

awesome then you just need to simplify [2sqrt(2)]^4 and 4*pi/4 and that's it

Answer 70

[2sqrt(2)]^4 =64
4*pi/4 = pi

Answer 71

good then the whole thing just becomes 64[cos(pi) + i*sin(pi)] and that's it