Question

Use W and Z to solve
Z= (-5sqrt3)/2+(5/2)i
W= 1+(sqrt3) i

A. Convert Z and W to polar form

Answer 1

\[r^2=x^2 +y^2 \implies r=\sqrt{x^2+y^2} \\ \tan(\theta)=\frac{y}{x} \\ \text{ do } \theta=\arctan(\frac{y}{x}) \text{ in first and forth quadrants } \\ \text{ do } \theta=\arctan(\frac{y}{x}) +\pi\text{ in second and third quadrants }\]

Answer 2

though you can add 2pi *n to those results depending on what range you want theta in

Answer 3

@freckles I think i figured it out by myself .... for Z at least

Converted to polar form it would be z=5(cos(5pi/6)+isin(5pi/6) right??

Answer 4

\[Z=\frac{-5 \sqrt{3}}{2}+\frac{5}{2}i \\ r=\sqrt{(\frac{-5 \sqrt{3}}{2})^2+(\frac{5}{2})^2}=\sqrt{\frac{25(3)}{4}+\frac{25}{4}}=\sqrt{\frac{25}{4}}\sqrt{3+1}=\frac{5}{2}\sqrt{4}=5\]
your r looks amazing

Answer 5

one sec checking your theta

Answer 6

ok :)

Answer 7

\[\theta=\arctan(\frac{\frac{5}{2}}{\frac{-5 \sqrt{3}}{2}}) +\pi \\ \theta=\arctan(\frac{5}{2} \cdot \frac{2 }{-5 \sqrt{3}})+ \pi \\ \theta=\arctan(\frac{-1}{\sqrt{3}})+\pi \\ \theta=\frac{-\pi}{6}+\pi\]
and your theta looks good as well

Answer 8

Ok, can you help me on W? I got pi/3 but i feel like that may not be right

Answer 9

so since W is in the first quadrant all that you need to do to calculate theta is do arctan(y/x)
y is sqrt(3)
and x=1

Answer 10

so your theta appears to be right

Answer 11

can you find r

Answer 12

Yay!! Can you help with a second question using these equations? It asks me to convert ZW using De Moivre Theorem but I dont know what de moivre theorem is

Answer 13

\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\]

Answer 14

so you multiply your r's
and add your thetas

Answer 15

when multiplying

Answer 16

if you wanted to do
\[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]

Answer 17

is that the form im supposed to use?

Answer 18

Well you asked to find AB not A/B
I was just sharing that last bit just in case you might find it useful later in the class

Answer 19

And I say you want to use the AB one because you said you want to find ZW

Answer 20

do you understand this:
\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\]
?

Answer 21

Yeah kind of, I just don't know where to plug in what

Answer 22

your r's for Z and W were 5 and 2 respectively
your theta's for Z and W were 5pi/6 and pi/3 respectively

your r_1=5 and r_2=2
and your theta_1=5pi/6 and your therta_2=pi/3

Answer 23

\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\]

\[ZW=5\cdot2(\cos(\frac{5\pi}{6}+\frac{\pi}{3})+i \sin(\frac{5\pi}{6}+\frac{\pi}{3}))\]

Answer 24

multiply the 5 and 2 there

and add the 5pi/6 and the pi/3 there

Answer 25

25 and 3.14??

Answer 26

5(2) is the same as saying 5+5 or 2+2+2+2+2
both of these should add up to be 10

Answer 27

ok hmm...if you aren't sure how to add 5pi/6 and pi/3
maybe just look at 5/6 and 1/3 and that then attach a pi

Answer 28

oooooh lol oops

Answer 29

\[\frac{5\pi}{6}+\frac{\pi}{3}=\pi(\frac{5}{6}+\frac{1}{3})=\pi(\frac{5}{6}+\frac{1}{3} \cdot \frac{2}{2}) \\ =\pi(\frac{5}{6}+\frac{2}{6})=?\]

Answer 30

pi(5/6+2/6) = 7pi/6??

Answer 31

yeah

Answer 32

\[ZW=2 \cdot 5 (\cos(\frac{5\pi}{6}+\frac{\pi}{3})+i \sin(\frac{5\pi}{6}+\frac{\pi}{3})) \\ ZW=10(\cos(\frac{7 \pi}{6})+i \sin(\frac{ 7\pi}{6}))\]

Answer 33

Thank you so much! Do you think you could help with one last part to this?? Its similar to what we just did so mybe ill have a better understanding

Answer 34

ok

Answer 35

C. Calculate z/w using de moivre theorem

Answer 36

ok well actually I mentioned that quotient above with A and B

Answer 37

let me copy and paste one sec

Answer 38

\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\]

\[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]

Answer 39

Identify your r1 and r2 and theta1 and theta2
and plug in the formula for A/B

Answer 40

where A=Z and B=W in this case

Answer 41

are the r's still 5 and 2?

Answer 42

yep because we are still using Z And W

Answer 43

and my thetas are 5pi/6 and pi/3 yes?

Answer 44

yep

Answer 45

(cos(5pi/6 -5pi/6) + isin(pi/3-pi/3) ??

Answer 46

for the inside you are suppose to do Z's theta - W's theta
and you still need Z's r / W's r on the outside of the ( )

Answer 47

\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\]

\[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]

do you how for A/B on the outside you have A's r divided by B's r
and the angles inside the cos and sin thing is A's theta- B's theta?

Answer 48

no im so confused right now lol

Answer 49

do you understand in that formula r1 is A's radius?

Answer 50

and that r2 is B's?

Answer 51

and that theta1 is A's angle thing
and theta2 is B's angle thing?

Answer 52

you are doing Z/W
so your r1 and theta1 will come from the information you got from Z
and r2 and theta2 will come from the information you got from W

Answer 53

earlier you told me r1=5 and r2=2
and theta1=5pi/6 and theta2=pi/3
just plug in the formula

Answer 54

\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\]

\[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]



\[Z=5(\cos(\frac{5 \pi}{6})+ i \sin(\frac{5 \pi}{6})) \\ W=2 (\cos(\frac{\pi}{3})+i \sin(\frac{\pi}{3})) \\ r_1=5,r_2=2 ,\theta_1=\frac{5\pi}{6}, \theta_2=\frac{\pi}{3} \\ \text{ plug into the formula } \\ \frac{Z}{W}=\frac{r_1}{r_2}(\cos(\theta_1-\theta_2)+i \sin(\theta_1-\theta_2))\]

Answer 55

replace the r1 with 5
replace the r2 with 2
theta1 with 5pi/6
theta 2 with pi/3

Answer 56

5(1)/2(2) = (cos(5pi/6-pi/3) +isin(5pi/6-pi/3) ??

Answer 57

ok but there shouldn't be an equal sign in between r1/r2 and (cos(thet...blah blah)
also where does the extra 1 and 2 come from?

Answer 58

oh wait i misunderstood its suppose to be 5/2

Answer 59

yep

Answer 60

and also remember to omit the equal sign

Answer 61

between the r1/r2 and the (cos...blah blah)

Answer 62

5/2(cos(5pi/6-pi/3) +isin(5pi/6-pi/3) ??

Answer 63

yep

Answer 64

is tht my answer

Answer 65

well just like we did 5pi/6+pi/3 earlier
you could do 5pi/6-pi/3

Answer 66

pi/2

Answer 67

\[\frac{5\pi}{6}-\frac{\pi}{3}=\pi(\frac{5}{6}-\frac{1}{3})=\pi(\frac{5}{6}-\frac{2}{6})=\pi(\frac{3}{6})=\pi \frac{1}{2}= \frac{\pi}{2}\]

Answer 68

oh yes

Answer 69

5/2 can be written as 2.5
but I think it looks just as pretty as 5/2