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

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freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[r^2=x^2 +y^2 \implies r=\sqrt{x^2+y^2} \\ \tan(\theta)=\frac{y}{x} \\ \text{ do } \theta=\arctan(\frac{y}{x}) \text{ in first and forth quadrants } \\ \text{ do } \theta=\arctan(\frac{y}{x}) +\pi\text{ in second and third quadrants }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2though you can add 2pi *n to those results depending on what range you want theta in

kalli
 one year ago
Best ResponseYou've already chosen the best response.0@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??

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[Z=\frac{5 \sqrt{3}}{2}+\frac{5}{2}i \\ r=\sqrt{(\frac{5 \sqrt{3}}{2})^2+(\frac{5}{2})^2}=\sqrt{\frac{25(3)}{4}+\frac{25}{4}}=\sqrt{\frac{25}{4}}\sqrt{3+1}=\frac{5}{2}\sqrt{4}=5\] your r looks amazing

freckles
 one year ago
Best ResponseYou've already chosen the best response.2one sec checking your theta

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\theta=\arctan(\frac{\frac{5}{2}}{\frac{5 \sqrt{3}}{2}}) +\pi \\ \theta=\arctan(\frac{5}{2} \cdot \frac{2 }{5 \sqrt{3}})+ \pi \\ \theta=\arctan(\frac{1}{\sqrt{3}})+\pi \\ \theta=\frac{\pi}{6}+\pi\] and your theta looks good as well

kalli
 one year ago
Best ResponseYou've already chosen the best response.0Ok, can you help me on W? I got pi/3 but i feel like that may not be right

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so 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
 one year ago
Best ResponseYou've already chosen the best response.2so your theta appears to be right

kalli
 one year ago
Best ResponseYou've already chosen the best response.0Yay!! 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2so you multiply your r's and add your thetas

freckles
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.0is that the form im supposed to use?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Well 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
 one year ago
Best ResponseYou've already chosen the best response.2And I say you want to use the AB one because you said you want to find ZW

freckles
 one year ago
Best ResponseYou've already chosen the best response.2do 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
 one year ago
Best ResponseYou've already chosen the best response.0Yeah kind of, I just don't know where to plug in what

freckles
 one year ago
Best ResponseYou've already chosen the best response.2your 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2)) \\ AB=r_1 r_2 (\cos(\theta_1+\theta_2)+i \sin(\theta_1+\theta_2))\] \[ZW=5\cdot2(\cos(\frac{5\pi}{6}+\frac{\pi}{3})+i \sin(\frac{5\pi}{6}+\frac{\pi}{3}))\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2multiply the 5 and 2 there and add the 5pi/6 and the pi/3 there

freckles
 one year ago
Best ResponseYou've already chosen the best response.25(2) is the same as saying 5+5 or 2+2+2+2+2 both of these should add up to be 10

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{5\pi}{6}+\frac{\pi}{3}=\pi(\frac{5}{6}+\frac{1}{3})=\pi(\frac{5}{6}+\frac{1}{3} \cdot \frac{2}{2}) \\ =\pi(\frac{5}{6}+\frac{2}{6})=?\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[ZW=2 \cdot 5 (\cos(\frac{5\pi}{6}+\frac{\pi}{3})+i \sin(\frac{5\pi}{6}+\frac{\pi}{3})) \\ ZW=10(\cos(\frac{7 \pi}{6})+i \sin(\frac{ 7\pi}{6}))\]

kalli
 one year ago
Best ResponseYou've already chosen the best response.0Thank 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

kalli
 one year ago
Best ResponseYou've already chosen the best response.0C. Calculate z/w using de moivre theorem

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok well actually I mentioned that quotient above with A and B

freckles
 one year ago
Best ResponseYou've already chosen the best response.2let me copy and paste one sec

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\] \[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1\theta_2)+i \sin(\theta_1\theta_2))\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2Identify your r1 and r2 and theta1 and theta2 and plug in the formula for A/B

freckles
 one year ago
Best ResponseYou've already chosen the best response.2where A=Z and B=W in this case

kalli
 one year ago
Best ResponseYou've already chosen the best response.0are the r's still 5 and 2?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yep because we are still using Z And W

kalli
 one year ago
Best ResponseYou've already chosen the best response.0and my thetas are 5pi/6 and pi/3 yes?

kalli
 one year ago
Best ResponseYou've already chosen the best response.0(cos(5pi/6 5pi/6) + isin(pi/3pi/3) ??

freckles
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\] \[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1\theta_2)+i \sin(\theta_1\theta_2))\] 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
 one year ago
Best ResponseYou've already chosen the best response.0no im so confused right now lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.2do you understand in that formula r1 is A's radius?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and that theta1 is A's angle thing and theta2 is B's angle thing?

freckles
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.2earlier you told me r1=5 and r2=2 and theta1=5pi/6 and theta2=pi/3 just plug in the formula

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\text{ Say} A=r_1(\cos(\theta_1)+i \sin(\theta_1)) \\ \text{ and } B=r_2(\cos(\theta_2)+i \sin(\theta_2))\] \[\frac{A}{B} \\ \text{ well that is } \frac{r_1}{r_2}(\cos(\theta_1\theta_2)+i \sin(\theta_1\theta_2))\] \[Z=5(\cos(\frac{5 \pi}{6})+ i \sin(\frac{5 \pi}{6})) \\ W=2 (\cos(\frac{\pi}{3})+i \sin(\frac{\pi}{3})) \\ r_1=5,r_2=2 ,\theta_1=\frac{5\pi}{6}, \theta_2=\frac{\pi}{3} \\ \text{ plug into the formula } \\ \frac{Z}{W}=\frac{r_1}{r_2}(\cos(\theta_1\theta_2)+i \sin(\theta_1\theta_2))\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2replace the r1 with 5 replace the r2 with 2 theta1 with 5pi/6 theta 2 with pi/3

kalli
 one year ago
Best ResponseYou've already chosen the best response.05(1)/2(2) = (cos(5pi/6pi/3) +isin(5pi/6pi/3) ??

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok 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
 one year ago
Best ResponseYou've already chosen the best response.0oh wait i misunderstood its suppose to be 5/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.2and also remember to omit the equal sign

freckles
 one year ago
Best ResponseYou've already chosen the best response.2between the r1/r2 and the (cos...blah blah)

kalli
 one year ago
Best ResponseYou've already chosen the best response.05/2(cos(5pi/6pi/3) +isin(5pi/6pi/3) ??

freckles
 one year ago
Best ResponseYou've already chosen the best response.2well just like we did 5pi/6+pi/3 earlier you could do 5pi/6pi/3

freckles
 one year ago
Best ResponseYou've already chosen the best response.2\[\frac{5\pi}{6}\frac{\pi}{3}=\pi(\frac{5}{6}\frac{1}{3})=\pi(\frac{5}{6}\frac{2}{6})=\pi(\frac{3}{6})=\pi \frac{1}{2}= \frac{\pi}{2}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.25/2 can be written as 2.5 but I think it looks just as pretty as 5/2
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