Question

?

Answer 1

first step would be to convert from rectangular to polar

Answer 2

\[5\sqrt{2}(\cos(-\frac{ \pi }{ 4 })+isin(-\frac{ \pi }{ 4 }))\]

Answer 3

may I ask how you got that theta value?

Answer 4

yes..

I used z=a+bi= absolute value of Z and then cos (theta)...

Answer 5

theta = arctan(y/x) = ?

Answer 6

wait x=5

Answer 7

IM alittle confused :/

Answer 8

remember that x is the first term and y is the second term (without the i part)

Answer 9

i think I got it.

Answer 10

I dont know if its right though :/

Answer 11

what did you get when you calculated arctan(y/x)?

Answer 12

NO i calculated all of it.

Answer 13

Can you take a look and see if i did it right

Answer 14

the theta value isn't -pi/4 and the r-value isn't 2

Answer 15

10^(1/5)[cos(pi/6 + 2pi n/6) + i sin(pi/5 + 2pi n/5)]
10^(1/5)[cos(pi/6) + i sin(pi/6)]
10^(1/5)[cos(pi/6 + 2pi/5) + i sin(pi/6 + 2pi/5)]
10^(1/5)[cos(pi/6 + 4pi/5) + i sin(pi/6 + 4pi/5)]
10^(1/5)[cos(pi/6 + 6pi/5) + i sin(pi/6 + 6pi/5)]
10^(1/5)[cos(pi/6 + 8pi/5) + i sin(pi/6 + 8pi/5)]

Answer 16

I know I redid it

Answer 17

pi/6 as the theta value? that can't be right

Answer 18

Hold on

Answer 19

10^(1/5)[cos(pi/6 + 2pi n/5) + i sin(pi/6 + 2pi n/5)] ----

10^(1/5)[cos(pi/6) + i sin(pi/6)]
10^(1/5)[cos(pi/6 + 2pi/5) + i sin(pi/6 + 2pi/5)]
10^(1/5)[cos(pi/6 + 4pi/5) + i sin(pi/6 + 4pi/5)]
10^(1/5)[cos(pi/6 + 6pi/5) + i sin(pi/6 + 6pi/5)] and
10^(1/5)[cos(pi/6 + 8pi/5) + i sin(pi/6 + 8pi/5)]

Answer 20

This maybe :/

Answer 21

can you show me how you got pi/6 as the angle? I just want to know where you might have messed up so I can correct your mistake

Answer 22

Okay

Answer 23

hold on one sec

Answer 24

r=1.64
tringular form

10^(1/5)[cos(pi/6) + i sin(pi/6)]

Answer 25

i went from polar to

Answer 26

the triangular

Answer 27

ugh i think Im doing this wrong :/

Answer 28

you just need to calculate arctan(y/x) where

x + yi is your original expression

making x = 5 and y = -5sqrt(3)

Answer 29

im getting a diff answer every time

Answer 30

-sqrt3

Answer 31

good and what is the arctan of that?

Answer 32

-pi/3

Answer 33

good, but since we're in the fourth quadrant, we need to add 2pi to that to get 5pi/3 as the angle theta

then we need to plug this into:

Answer 34

where r is the modulus (10), n is the root number (since want the 5th root we need n = 5), and k is the set of integers from 0,1,2,3,4 which will give you five total roots after all the k values have been calculated

Answer 35

\[\sqrt[5]{10}(\cos \frac{ 5\pi/3 +2\pi(k)}{ 5 }+isin \frac{ 5\pi/3+2\pi(k) }{ 5 })\]

Answer 36

good, then start plugging in k = 0 ,1,2,3, and 4

Answer 37

Been like that forever

Answer 38

\[\sqrt[5]{10}(\cos \frac{ 5\pi/3 +2\pi(k)}{ 5 }+isin \frac{ 5\pi/3+2\pi(k) }{ 5 })\]

you just need to copy this and replace k with the appropriate value

Answer 39

put dont i have to solve it or is it just the equation

Answer 40

you just need the equation

Answer 41

5sqrt10 (cos (5pi/3) + 2pi (0)/5 + isin((5pi/3)+2pi(0)/2)
5sqrt10 (cos (5pi/3) + 2pi (1)/5 + isin((5pi/3)+2pi(1)/2)
5sqrt10 (cos (5pi/3) + 2pi (2)/5 + isin((5pi/3)+2pi(2)/2)
5sqrt10 (cos (5pi/3) + 2pi (3)/5 + isin((5pi/3)+2pi(3)/2)
5sqrt10 (cos (5pi/3) + 2pi (4)/5 + isin((5pi/3)+2pi(4)/2)
5sqrt10 (cos (5pi/3) + 2pi (5)/5 + isin((5pi/3)+2pi(5)/2)

Answer 42

then you just need a bit of algebra to simplify the expressions inside cos and sin and then you'll be done

Answer 43

like, you would just simplify [ 5pi/3 + 2pi*0 ] / 5, etc. for each root, then you're done

Answer 44

[ 5pi/3 + 2pi*0 ] / 5
[ 5pi/3 + 2pi*1 ] / 5
[ 5pi/3 + 2pi*2 ] / 5
[ 5pi/3 + 2pi*3 ] / 5
[ 5pi/3 + 2pi*4 ] / 5
[ 5pi/3 + 2pi*5 ] / 5

Answer 45

like that

Answer 46

good but you would keep simplifying

Answer 47

like what does [ 5pi/3 + 2pi*0 ] / 5 = in simpler terms?

Answer 48

notice how 2*pi*0 = 0 so you can just write this as (5pi/3) / 5 = pi/3

making the first root (10)^(1/5) * [cos(pi/3) + isin(pi/3)]

then repeat this process for the other four k values

Answer 49

(5pi/3) / 4 = 5pi/12
(5pi/3) / 3 = 5pi/9
(5pi/3) / 2 = 5pi/6
(5pi/3) / 1 = 5pi/3
(5pi/3) / 0 = -infinity

(10)^(1/5) * [cos(5pi/12) + isin(5pi/12)]
(10)^(1/5) * [cos(5pi/9) + isin(5pi/9)]
(10)^(1/5) * [cos(5pi/6) + isin(5pi/6)]
(10)^(1/5) * [cos(5pi/3) + isin(5pi/3)]
(10)^(1/5) * [cos(-infinity) + isin(-infinity)]

Answer 50

the number in the denominator is not the one that changes. the denominator is always 5.

[ 5pi/3 + 2pi*k ] / 5

notice how k is in the numerator

Answer 51

(5pi/2) / 5 =

Answer 52

like this

Answer 53

start with [ 5pi/3 + 2pi*k ] / 5

we already did k = 0 so let's move on to k = 1

replace "k" with 1

[ 5pi/3 + 2pi*1 ] / 5 = ?

simplify this, then go back to the root expression and plug this in for cos and sin

Answer 54

11pi/3

Answer 55

still need to divide by 5 at the end, it then becomes 11pi/15

then plug this in to the root to get 10^(1/5) * [cos(11pi/15) + i*sin(11pi/15)]

then keep going for k = 2,3,4

Answer 56

[ 5pi/3 + 2pi*2 ] / 5 = 17pi/3

=17pi/15
10^(1/5) * [cos(17pi/15) + i*sin(17pi/15)]

Answer 57

like that

Answer 58

good

Answer 59

uh thank god. im gonna do the rest now

Answer 60

denominator should be 15 not 3

Answer 61

10^(1/5) * [cos(23pi/15) + i*sin(23pi/15)]
10^(1/5) * [cos(29pi/15) + i*sin(29pi/15)]

Answer 62

=)

Answer 63

awesome, so those are your five roots and that's it