anonymous
  • anonymous
compute the modulus and argument of each complex number. a) 1+i
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@mathmate could you help with this one?
zepdrix
  • zepdrix
Eyyy :) So what are we having trouble with? The modulus is just the length directly to that point, ya? So like,|dw:1433125698294:dw|
zepdrix
  • zepdrix
|dw:1433125748356:dw|So we have this going on, ya? :)

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zepdrix
  • zepdrix
Mmm sorry sloppy hand writing -_- that's a \(\Large\rm |z|\) on the graph
zepdrix
  • zepdrix
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
anonymous
  • anonymous
I got sqrt 2 for my modulus and pi/4 for my argument
anonymous
  • anonymous
zepdrix
  • zepdrix
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 >.<
zepdrix
  • zepdrix
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 :)
zepdrix
  • zepdrix
9pi/4, not 5pi/4, woops* bah anyway
anonymous
  • anonymous
You are right! Thank you!

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