anonymous
  • anonymous
express in polar form r cisθ for: a) -1-i b) -3+4i (to 2 decimal places of a degree) c) use your answers to parts a) and b) to simplify (-3+4i)/(-1-i)^2 in polar form.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
first you need \[r=\sqrt{a^2+b^2}\] so for part a you get \[r=\sqrt{2}\] and for part b you get \[r=5\]
anonymous
  • anonymous
then you need \[\theta\] which you get by solving \[tan(\frac{b}{a})=\theta\]
anonymous
  • anonymous
are you using degree or radians for this? if radians the first one is \[\frac{5\pi}{4}\] if degrees is it 225

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anonymous
  • anonymous
if you plot -1-i you see you are in quadrant III which is why you get 225
anonymous
  • anonymous
for part b you get \[tan^{-1}(-\frac{4}{3})\] from a calculator then you have to add 180 to your answer. let me see if i have one
anonymous
  • anonymous
i get 126.78 to two places but you should check it
anonymous
  • anonymous
so answer to first one is \[\sqrt{2}(cos(45)+isin(45))\] and second is \[5(cos(126.78)+isin(126.78))\]
anonymous
  • anonymous
then to divide, you divide the r's and subtract the angles.
anonymous
  • anonymous
what does it say next to (45)? its all blurry. thanks for your help so far!
anonymous
  • anonymous
ok first of all i made a mistake. it is not 45 it should be 225
anonymous
  • anonymous
i don't know why i wrote 45 because earlier i said the angle is 225 degrees
anonymous
  • anonymous
it says cos(225)+isin(225)
anonymous
  • anonymous
i guess the latex is hard to read but may be clearer if you refresh the browser, that often works for me
anonymous
  • anonymous
oh thanks for the tip!
anonymous
  • anonymous
your next job is to square \[(-1-i)\] which is actually very easy not in polar form, but i guess they want it in polar form so we do it
anonymous
  • anonymous
you do it by squaring r and doubling the angle so you get \[(-1-i)^2=\sqrt{2}^2(cos(225\times 2)+isin(225\times2))\] \[=2cos(450)+isin(450))\] \[=2(cos(90)+isin(90))\]
anonymous
  • anonymous
the last equality because sine and cosine are periodic with period 360 so i just subtracted 360 from 450 to get 90
anonymous
  • anonymous
btw this problem is really stupid because \[(-1-i)^2\] is just 2i and dividing by 2i is easy. very annoying to put in polar form. but in any case we see that since cos(90)=0 and sin(90)=1 we get \[(-1-i)^2=2i\] using polar form
anonymous
  • anonymous
your last job is to divide. as i said dividing by 2i is easy but they want you to use polar form so you divide 5 by 2 and subtract 90 from 126.78 to get the answer of \[\frac{5}{2}(cos(36.78)+isin(36.78))\]
anonymous
  • anonymous
now the snap way: \[\frac{-3+4i}{(-1-i)^2}=\frac{-3+4i}{2i}\] \[=\frac{-3+4i}{2i}\times \frac{-i}{-i}=\frac{4+3i}{2}=2+\frac{3}{2}i\]\]
anonymous
  • anonymous
thank you that answer matches the answer in the book :)
anonymous
  • anonymous
good. hope you understood the steps and that dividing by 2i is much easier without converting to polar for. easy to divide!

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