I cannot solve two problems regarding complex equations.

1)Let $z^2+w^2=0$, prove that $$z^{4n+2}+w^{4n+2}=0, n \in \mathbb{N^{*}}$$

What I tried; $$z^2 \cdot z^{4n}+w^2 \cdot w^{4n}=0 \iff w^2(w^{4n}-z^{4n})=0$$ but it doesn't really prove anything.

2) Let $z=\frac{1+i\sqrt{3}}{2}$, evaluate $1+z^{1997}-z^{1998}+z^{1999}$

I can think of a way turning $z=e^{\frac{i\pi}{3}}$ but I have to solve it without anything but complex number properties(and algebra of course).




See Proof of $a^n+b^n$ divisible by a+b when n is odd


$$z=\frac{1+i\sqrt{3}}{2}\iff 2z-1=i\sqrt3$$

Squaring we get, $$4z^2-4z+1=-3\iff z^2-z+1=0$$

$$\implies z^{1997}-z^{1998}+z^{1999}=z^{1997}(1-z+z^2)=0$$

  • $\begingroup$ @user148432, How about this $\endgroup$ – lab bhattacharjee Aug 29 '14 at 3:30
  • $\begingroup$ This is much easier, thanks. $\endgroup$ – UserX Aug 29 '14 at 6:00

For the first one isn't it valid to say:$$z^2=-w^2$$$$\therefore z^4=w^4$$$$\therefore z^{4n}=w^{4n}$$$$\therefore z^{4n+2}+w^{4n+2}=z^2\cdot z^{4n}+w^2\cdot w^{4n}$$$$=-w^2\cdot z^{4n}+w^2\cdot w^{4n}=w^2(-z^{4n}+w^{4n})=0$$

  • $\begingroup$ That's what I did. But how does that prove it? $\endgroup$ – UserX Aug 28 '14 at 21:36
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    $\begingroup$ I cannot see how it doesn't prove it? $\endgroup$ – Mufasa Aug 28 '14 at 21:37
  • $\begingroup$ Now I saw the part where $z^4=w^4$, now it does. Thanks. What about the second one? $\endgroup$ – UserX Aug 28 '14 at 21:38
  • $\begingroup$ I think @amcalde has given enough of a hint at the proof of the second one in his last comment (i.e. make use of the fact that $z^3=-1$). $\endgroup$ – Mufasa Aug 28 '14 at 21:39
  • $\begingroup$ I solved both, thanks a lot guys :D $\endgroup$ – UserX Aug 28 '14 at 21:44

1) Rearrange your equation, conversely just plug in $z^2 = -w^2$.

2) You are right, use your $z = e^{i\frac{\pi}{3}}$

You just need to think about what the exponents are mod 6. That's all that matters here. Do you see why?

  • $\begingroup$ That's exactly what I did in 1) and in 2) i stated I can't use complex exponentials... And yes I do understand why, I have to find the correct full $2\pi$ revolutions plus $c$, and my $z$ is $e^c$. But I cannot use that method. $\endgroup$ – UserX Aug 28 '14 at 21:28
  • $\begingroup$ OK I was unclear on that I guess. But then you just need to prove that $z^6 == 1$ (as written without exponentials). The answer is still valid. Finding, say, 1998 mod 6 is still all you need to do essentially. (It's actually easier to see that $z$ cubes to -1). $\endgroup$ – amcalde Aug 28 '14 at 21:30
  • $\begingroup$ I cannot use mod either... $\endgroup$ – UserX Aug 28 '14 at 21:32
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    $\begingroup$ OK but since $z^3 = -1$ then you just say something like $z^{1998} = (z^3)^{666} = (-1)^{666} = 1^{333} = 1$ $\endgroup$ – amcalde Aug 28 '14 at 21:35

It's rather simple. $$z^2+w^2=0$$ Then $$z^2=-w^2$$ Which means, $$z^4=w^4$$ and for all $n$ : $$z^{4n}=w^{4n}$$ Therefore $$4z^{4n}z^2=-w^{4n}w^2$$ as a result, for all $n$, $$z^{4n+2}+w^{4n+2}=0$$


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