anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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jtvatsim
  • jtvatsim
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|
jtvatsim
  • jtvatsim
This will always be unique as long as we don't "wrap around" more than once with our angles. Like this: |dw:1439528634149:dw|
jtvatsim
  • jtvatsim
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.

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anonymous
  • anonymous
okay I think I follow
jtvatsim
  • jtvatsim
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. :)
jtvatsim
  • jtvatsim
But hold that thought one moment, I do have to check one thing before... I may have made a mistake for 0. ... thinking...
anonymous
  • anonymous
So it's true or?
jtvatsim
  • jtvatsim
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?
jtvatsim
  • jtvatsim
@ganeshie8
anonymous
  • anonymous
Thanks for your help anyway! I'll let you know what I get:)
jtvatsim
  • jtvatsim
Alright, please do. Good luck!
anonymous
  • anonymous
@jtvatsim I got it right, It was True:)
jtvatsim
  • jtvatsim
Phew... close one! :)

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