Complex number equations

Question

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).

score 3 · Answer 1 · answered Aug 28 '14 at 21:33

3

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

answered Aug 28 '14 at 21:33

Mufasa

5,434

That's what I did. But how does that prove it? – UserX Aug 28 '14 at 21:36
1

I cannot see how it doesn't prove it? – Mufasa Aug 28 '14 at 21:37
Now I saw the part where $z^4=w^4$, now it does. Thanks. What about the second one? – UserX Aug 28 '14 at 21:38
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 $z^3=-1$). – Mufasa Aug 28 '14 at 21:39
I solved both, thanks a lot guys :D – UserX Aug 28 '14 at 21:44
@user148432 - yw :) – Mufasa Aug 28 '14 at 21:45

score 3 · Accepted Answer · edited Apr 13 '17 at 12:21

3

$(1)$

$$z^{4n+2}+w^{4n+2}=z^{4n+2}+(w^2)^{2n+1}=z^{4n+2}+(-z^2)^{2n+1}=z^{4n+2}-z^{4n+2}=0$$

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

$(2)$

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

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

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

edited Apr 13 '17 at 12:21

Community

1

answered Aug 29 '14 at 03:29

lab bhattacharjee

274,582

@user148432, How about this – lab bhattacharjee Aug 29 '14 at 03:30
This is much easier, thanks. – UserX Aug 29 '14 at 06:00

score 2 · Answer 3 · answered Aug 28 '14 at 21:25

2

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?

answered Aug 28 '14 at 21:25

amcalde

4,674

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. – UserX Aug 28 '14 at 21:28
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). – amcalde Aug 28 '14 at 21:30
I cannot use mod either... – UserX Aug 28 '14 at 21:32
1

OK but since $z^3 = -1$ then you just say something like $z^{1998} = (z^3)^{666} = (-1)^{666} = 1^{333} = 1$ – amcalde Aug 28 '14 at 21:35

score 0 · Answer 4 · answered Aug 28 '14 at 21:57

0

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 $n$, $$z^{4n+2}+w^{4n+2}=0$$

answered Aug 28 '14 at 21:57

Complex number equations

4 Answers4