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).

Answer 1

$(1)$

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

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

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

Answer 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?

Answer 3

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

Answer 4

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