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stutiRastogi

This representation would take care of the full circle rotation, right? (i.e., when we go past 2$\pi $) Because say the sum goes to 3$\pi $, we get $$e^{\iota3\pi} = e^{\iota\pi} * e^{\iota\pi} * e^{\iota\pi} = -1 * -1 * -1 = -1$$

So, we don't need to take care of modulus or anything. Is this the answer to the question?

keenan

Yes! Exactly. The convenience of this representation is that we never need to think or worry about whether two angles are "the same up to some multiple of 2$pi$". Essentially, we just have a unique complex number representing each angle between 0 and 2$pi$.