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anonymous

  • one year ago

For a complex number written in polar form, its representation is unique as long as the argument is restricted to [0,2π). True or false?

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  1. jtvatsim
    • one year ago
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    This one is a little tricky, but not too bad.The way polar form locates complex numbers is by giving you a radius and an angle like this:|dw:1439528539759:dw|

  2. jtvatsim
    • one year ago
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    This will always be unique as long as we don't "wrap around" more than once with our angles. Like this: |dw:1439528634149:dw|

  3. jtvatsim
    • one year ago
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    The question says that the angle is restricted to [0, 2*pi) and since 2*pi is 360 degrees, this means that the angle won't "wrap around." The answer is True.

  4. anonymous
    • one year ago
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    okay I think I follow

  5. jtvatsim
    • one year ago
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    Good! Hopefully you see the idea behind the question. As long as we restrict to [0, 2*pi) we will always have a unique representation. We won't have "weird things" like z = cos(2*pi) = cos(4*pi) = cos(6*pi), etc. Each number will have just one value. In other words, it will be unique. :)

  6. jtvatsim
    • one year ago
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    But hold that thought one moment, I do have to check one thing before... I may have made a mistake for 0. ... thinking...

  7. anonymous
    • one year ago
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    So it's true or?

  8. jtvatsim
    • one year ago
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    Well, there is still the weird case that you could have z = r(cos(theta) + i sin(theta)) the polar form and yet, 0 = 0(cos(theta) + i sin(theta)) for any angle you could pick so it's not unique... but then, some textbooks just say that 0 will always have an angle of 0... ARGH! lol ... it's a tough call. If we ignore 0, then the answer is True. If it's a trick question and we are supposed to look out for 0, then the answer is False. Maybe someone else has a different insight?

  9. jtvatsim
    • one year ago
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    @ganeshie8

  10. anonymous
    • one year ago
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    Thanks for your help anyway! I'll let you know what I get:)

  11. jtvatsim
    • one year ago
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    Alright, please do. Good luck!

  12. anonymous
    • one year ago
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    @jtvatsim I got it right, It was True:)

  13. jtvatsim
    • one year ago
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    Phew... close one! :)

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