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anonymous

  • 5 years ago

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.

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  1. anonymous
    • 5 years ago
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    first you need \[r=\sqrt{a^2+b^2}\] so for part a you get \[r=\sqrt{2}\] and for part b you get \[r=5\]

  2. anonymous
    • 5 years ago
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    then you need \[\theta\] which you get by solving \[tan(\frac{b}{a})=\theta\]

  3. anonymous
    • 5 years ago
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    are you using degree or radians for this? if radians the first one is \[\frac{5\pi}{4}\] if degrees is it 225

  4. anonymous
    • 5 years ago
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    if you plot -1-i you see you are in quadrant III which is why you get 225

  5. anonymous
    • 5 years ago
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    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

  6. anonymous
    • 5 years ago
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    i get 126.78 to two places but you should check it

  7. anonymous
    • 5 years ago
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    so answer to first one is \[\sqrt{2}(cos(45)+isin(45))\] and second is \[5(cos(126.78)+isin(126.78))\]

  8. anonymous
    • 5 years ago
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    then to divide, you divide the r's and subtract the angles.

  9. anonymous
    • 5 years ago
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    what does it say next to (45)? its all blurry. thanks for your help so far!

  10. anonymous
    • 5 years ago
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    ok first of all i made a mistake. it is not 45 it should be 225

  11. anonymous
    • 5 years ago
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    i don't know why i wrote 45 because earlier i said the angle is 225 degrees

  12. anonymous
    • 5 years ago
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    it says cos(225)+isin(225)

  13. anonymous
    • 5 years ago
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    i guess the latex is hard to read but may be clearer if you refresh the browser, that often works for me

  14. anonymous
    • 5 years ago
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    oh thanks for the tip!

  15. anonymous
    • 5 years ago
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    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

  16. anonymous
    • 5 years ago
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    you do it by squaring r and doubling the angle so you get \[(-1-i)^2=\sqrt{2}^2(cos(225\times 2)+isin(225\times2))\] \[=2cos(450)+isin(450))\] \[=2(cos(90)+isin(90))\]

  17. anonymous
    • 5 years ago
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    the last equality because sine and cosine are periodic with period 360 so i just subtracted 360 from 450 to get 90

  18. anonymous
    • 5 years ago
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    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

  19. anonymous
    • 5 years ago
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    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 \[\frac{5}{2}(cos(36.78)+isin(36.78))\]

  20. anonymous
    • 5 years ago
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    now the snap way: \[\frac{-3+4i}{(-1-i)^2}=\frac{-3+4i}{2i}\] \[=\frac{-3+4i}{2i}\times \frac{-i}{-i}=\frac{4+3i}{2}=2+\frac{3}{2}i\]\]

  21. anonymous
    • 5 years ago
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    thank you that answer matches the answer in the book :)

  22. anonymous
    • 5 years ago
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    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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