krypticlemonjuice
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- HSC
- 2023
Can someone please explain the solution. How did they get those values for arg and re of w and w^2.
Also why did they group w and w^5, and w^2 and w^4, and so on...
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Neap Q (1 Viewer)
- Thread starter krypticlemonjuice
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Can someone please explain the solution. How did they get those values for arg and re of w and w^2.
Also why did they group w and w^5, and w^2 and w^4, and so on...
Because w^6=1, w is a sixth root of unity. Hence, you can just divide 2π by 6 to get the argument. From there, to be able to be powered into w^6, the modulus of w must also be 1. Therefore, w is just 1cis(π/3). To find w^2, we just double the argument and keep the modulus the same, and likewise we triple the argument for w^3 and so on. If you map out all of the roots, you’ll see that w and w^5 are the conjugate of one another (makes sense as they are both one “w” away from w^6), so when they add the imaginary parts cancel out, and the identical real parts add to each other, similar story for w^2 and w^4. Graphing these questions helps a lot, I highly recommend it.