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anonymous

  • one year ago

compute the modulus and argument of each complex number. a) 1+i

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  1. anonymous
    • one year ago
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    @mathmate could you help with this one?

  2. zepdrix
    • one year ago
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    Eyyy :) So what are we having trouble with? The modulus is just the length directly to that point, ya? So like,|dw:1433125698294:dw|

  3. zepdrix
    • one year ago
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    |dw:1433125748356:dw|So we have this going on, ya? :)

  4. zepdrix
    • one year ago
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    Mmm sorry sloppy hand writing -_- that's a \(\Large\rm |z|\) on the graph

  5. zepdrix
    • one year ago
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    Or if you're more comfortable looking at it in polar form, you can do the steps to convert it into \(\Large\rm z=r e^{i\theta}\) Where our modulus is r, and argument is the angle theta, err principle argument at least. Mmmmm what are you thinking ms Hailey? :d

  6. anonymous
    • one year ago
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    I got sqrt 2 for my modulus and pi/4 for my argument

  7. anonymous
    • one year ago
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    @zepdrix

  8. zepdrix
    • one year ago
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    yayyyy good job \c:/ do they just want pi/2 for the arg z? Or do they want the rotations and stuff also? so maybe pi/2 + 2k pi. I dunno i can't remember >.<

  9. zepdrix
    • one year ago
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    The `principle` argument, yes, is \(\Large\rm \frac{\pi}{4}\). But notice that \(\Large\rm \frac{5\pi}{4}\) is also an acceptable angle, I just spun a full time around the circle and landed in the same spot. If you haven't talked about that in your class, maybe we shouldn't worry about it :)

  10. zepdrix
    • one year ago
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    9pi/4, not 5pi/4, woops* bah anyway

  11. anonymous
    • one year ago
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    You are right! Thank you!

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