Use W and Z to solve
Z= (-5sqrt3)/2+(5/2)i
W= 1+(sqrt3) i
A. Convert Z and W to polar form

- kalli

Use W and Z to solve
Z= (-5sqrt3)/2+(5/2)i
W= 1+(sqrt3) i
A. Convert Z and W to polar form

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- freckles

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

- freckles

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

- kalli

@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??

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- freckles

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

- freckles

one sec checking your theta

- kalli

ok :)

- freckles

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

- kalli

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

- freckles

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

- freckles

so your theta appears to be right

- freckles

can you find r

- kalli

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

- freckles

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

- freckles

so you multiply your r's
and add your thetas

- freckles

when multiplying

- freckles

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

- kalli

is that the form im supposed to use?

- freckles

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

- freckles

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

- freckles

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

- kalli

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

- freckles

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

- freckles

\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\]
\[ZW=5\cdot2(\cos(\frac{5\pi}{6}+\frac{\pi}{3})+i \sin(\frac{5\pi}{6}+\frac{\pi}{3}))\]

- freckles

multiply the 5 and 2 there
and add the 5pi/6 and the pi/3 there

- kalli

25 and 3.14??

- freckles

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

- freckles

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

- kalli

oooooh lol oops

- freckles

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

- kalli

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

- freckles

yeah

- freckles

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

- kalli

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

- freckles

ok

- kalli

C. Calculate z/w using de moivre theorem

- freckles

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

- freckles

let me copy and paste one sec

- freckles

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

- freckles

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

- freckles

where A=Z and B=W in this case

- kalli

are the r's still 5 and 2?

- freckles

yep because we are still using Z And W

- kalli

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

- freckles

yep

- kalli

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

- freckles

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 ( )

- freckles

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

- kalli

no im so confused right now lol

- freckles

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

- freckles

and that r2 is B's?

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- kalli

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

- freckles

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?

- kalli

oh wait i misunderstood its suppose to be 5/2

- freckles

yep

- freckles

and also remember to omit the equal sign

- freckles

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

- kalli

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

- freckles

yep

- kalli

is tht my answer

- freckles

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

- kalli

pi/2

- freckles

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

- freckles

oh yes

- freckles

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.