anonymous
  • anonymous
11. Use w and z to solve. z = (- 5 √3 )/ 2 + (5/2)i w = 1 + (√3) i a. Convert z and w to polar form. b. Convert zw using De Moivre’s Theorem. c. Calculate z / w using De Moivre’s Theorem.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
mathslover
  • mathslover
First of all. We shall solve a) i) Converting z into polar form : If a + bi is the complex number then \(r\cos \theta = a\) and \(r\sin \theta = b\) Similarly, we have here : \(z = \cfrac{-5\sqrt{3}}{2} + \cfrac{5}{2} i \) as the complex number. Can you tell me what is \(r\cos \theta\) and \(r \sin \theta \) here ... ?
anonymous
  • anonymous
juicy question. i'll leave this to u bro
anonymous
  • anonymous
im not sure

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mathslover
  • mathslover
You there @kaylalynn ?
anonymous
  • anonymous
ya
anonymous
  • anonymous
i said im not sure. :(
mathslover
  • mathslover
Oh;. I didn't notice that.
mathslover
  • mathslover
|dw:1371454397613:dw|
mathslover
  • mathslover
Now compare : a+bi and the given complex number (z = \(\cfrac{-5\sqrt{3}}{2} + \cfrac{5}{2} i \) ) .
mathslover
  • mathslover
So, we get by comparing : a = \(\cfrac{-5\sqrt{3}}{2} \) and b = \(\cfrac{5}{2} \) Therefore : \(\cfrac{-5\sqrt{3}}{2} = r\cos \theta\) and \(\cfrac{5}{2} = r\sin \theta\)
anonymous
  • anonymous
It is very simple : The polar form of a nonzero complex number z=a+bi is : \[z=r(\cos\theta+i\sin\theta)\] Where : \[r=|z|=\sqrt{a^2+b^2}\\ \cos\theta=\frac a{|z|}\\ \sin\theta=\frac{b}{|z|}\]
mathslover
  • mathslover
Are you getting my point now @kaylalynn ?
anonymous
  • anonymous
i think so
mathslover
  • mathslover
\(r\cos \theta = \cfrac{-5\sqrt{3}}{2} \) \(r\sin \theta = \cfrac{5}{2} \) Square both equations : \(r^2\cos^2 \theta = \cfrac{75}{4}\) \(r^2 \sin^2 \theta = \cfrac{25}{4} \) Add both now : \(r^2 \cos^2 \theta + r^2 \sin^2 \theta = \cfrac{75}{4} + \cfrac{25}{4} \)
mathslover
  • mathslover
Take : \(r^2\) common ,solve for "r" there first.
mathslover
  • mathslover
Can you do that @kaylalynn ?
anonymous
  • anonymous
100/4?
mathslover
  • mathslover
Yes, but that is : r^2 ... Square root both sides
mathslover
  • mathslover
You have : \(r^2 = \cfrac{100}{4} \implies r^2 = 25\) , right?
anonymous
  • anonymous
ya
mathslover
  • mathslover
Square root both sides to solve for "r" \(\sqrt{r^2} = \sqrt{25}\) \(r = 5\)
mathslover
  • mathslover
Now, as \(r\cos \theta = \cfrac{-5\sqrt{3}}{2} \) Put r =5 in the equation and solve for theta.
mathslover
  • mathslover
Can you tell me what you get for theta?
anonymous
  • anonymous
how would i do that?
mathslover
  • mathslover
See , we have : \(5\cos \theta = \cfrac{-5\sqrt{3}}{2} \) , right?
anonymous
  • anonymous
So, for example : \[z=\frac{- 5 \sqrt3 } 2 + \frac52i\] So : \[r=|z|=\sqrt{\left(\frac{- 5 \sqrt3 } 2\right)^2+\left(\frac52\right)^2}=\sqrt{\frac{3\times25+25}{4}}=\frac{10}{2}=5\] And then : \[\cos\theta=\frac{a}{|z|}=\frac{\frac{- 5 \sqrt3 } 2}{5}=\frac{- \sqrt3 } 2\\ \sin\theta=\frac{b}{|z|}=\frac{\frac52}{5}=\frac12 \] hence : \[\theta=\pi-\frac\pi6=\frac{5\pi}{6}\] Therefore, the polar form of z is : \[z=5(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})\]
mathslover
  • mathslover
@Noura11, it is better to interact with the asker, rather than doing all the things in a single post. Please refer os coc and http://openstudy.com/study#/updates/51badc02e4b066f3d785f98d
mathslover
  • mathslover
Kaylalynn , where are you having problem in solving for theta?
anonymous
  • anonymous
i was having trouble isolating it because having the division divided looked weird and i wasnt sure if it was suppose to be looking like that.
mathslover
  • mathslover
Division will not work here. cos theta is a function... Divide both sides by 5 ... what do you get there ?
mathslover
  • mathslover
Or, first solve for "cos theta"
mathslover
  • mathslover
|dw:1371456119369:dw|
anonymous
  • anonymous
right
mathslover
  • mathslover
Now I can write that as : |dw:1371456624926:dw|
mathslover
  • mathslover
Can you tell me when does cos theta equal -root3/2 ?
mathslover
  • mathslover
|dw:1371456690755:dw|
mathslover
  • mathslover
|dw:1371456732275:dw| Did you get it?

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