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Considering the complex number $j$ such that $$ j = \frac{-1}{2} + i\frac{\sqrt3}{2} $$ Prove that $ \forall n \in \Bbb Z : $ $$ ( j^{2n} - j^n ) \in i\Bbb R $$

( $i\Bbb R$ being the set of pure imaginary numbers)

Guy Fsone
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Abbkey
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  • @Useless I considered $ u = j^{2n} - j^n $ and tried to prove that its complex conjugate was equal to $-u$ but that was to no avail – Abbkey Jan 19 '18 at 12:21
  • Why the upvotes? This is pure do-my-hwk-for-me-asap stuff. – Did Jan 27 '18 at 12:38
  • Shouldn't I upvote the answers that were helpful ? I don't understand what you mean – Abbkey Jan 27 '18 at 12:46
  • You should first and foremost start posting questions with context. That the present question received 4 upvotes signals a serious misunderstanding of the purpose of this site. – Did Jan 27 '18 at 13:34
  • Sure. I'll keep that in mind if I post any new question. Thanks for the clarification. – Abbkey Jan 27 '18 at 13:40

3 Answers3

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Observing that $$j^2=\left(\frac{-1}{2} + i\frac{\sqrt3}{2}\right)^2=-\frac{1}{2} - i\frac{\sqrt3}{2}=\overline{j}$$

then since $\Im(z) =\frac{1}{2i}(z-\overline{z})$ we obviously get $$j^{2n}-j^n =\bar{j}^{n}-j^n =\overline{j^n} - j^n = \color{blue}{-2i\Im(j^n) \in i\Bbb R} $$

Guy Fsone
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  • This is exactly how I ended up proving it – Abbkey Jan 19 '18 at 18:20
  • @Abbkey Why did you not edit your answer here it could be helpful – Guy Fsone Jan 19 '18 at 18:22
  • I added your answer since it's similar to mine and would save me a lot of typing time, if you don't mind of course – Abbkey Jan 19 '18 at 18:30
  • @Abbkey Also don't forget to put a green-checkmark (see it the right to the answers) on your favored answer it is important for the visitors and reader of your post and to the incoming users of MSE – Guy Fsone Jan 19 '18 at 18:32
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Hint:

Note that using Euler's formula, we can write $$ j = \cos \frac{2\pi}3 + i\sin \frac{2\pi}3 = e^{\dfrac{2\pi i} 3}$$ which is a complex root of unity.

  • @Abbkey Even without Euler's, you may notice that $j$ is a complex root of unity, so: $$j^2 = \bar j$$ the conjugate of $j$. Thus, in a few steps we can easily show that $$j^{2n}-j^n = \bar j^n - j^n \in i\mathbb R$$ –  Jan 19 '18 at 12:33
  • I am not sure whether your $\overline j^n-j^n$ was meant to represent $\left(\bar j\right)^n-j^n$ or $\overline {\left(j^n\right)}-j^n$ - they are of course the same thing, though the second is more obviously purely imaginary – Henry Jan 19 '18 at 12:39
  • @Rohan that definitely helps. Thank you – Abbkey Jan 19 '18 at 12:44
  • Why the downvote on mine and @Lab's answers? –  Jan 20 '18 at 08:18
  • @Downvoter: I see you have downvoted my answers in a serial fashion. I don't care about it and may God judge you. –  Jan 20 '18 at 08:22
  • There must be a misunderstanding here. I clicked on the arrow pointing up, because I intuitively though that that was how you upvote the answers. I don't see why I would downvote since you're trying to help me. Maybe there something wrong with the app but again I didn't downvote. – Abbkey Jan 21 '18 at 12:38
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$$j=e^{2\pi i/3}$$

$$j^m=e^{2\pi i m/3}=\cos\dfrac{2m\pi }3+i\sin\dfrac{2m\pi }3$$ using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i*\sin(\varphi)$?

Now the real part $j^{2n}-j^n$

$$=\cos\dfrac{4n\pi }3-\cos\dfrac{2n\pi}3$$ $$=-2\sin2n\pi\sin\dfrac{2n\pi}3=?$$

lab bhattacharjee
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  • Can u suggest another way to proving it since i'm not allowed to use Euler's formula, we haven't studied that just yet. – Abbkey Jan 19 '18 at 12:28