Let $z \in \mathbb C$, such that $z = x+ix, \; \forall x \in \mathbb R^* $
Prove that $$K(z) = \frac {z^4 + z^8 + \cdots+ z^{4n}} {iz^2 + i^5z^6 + \cdots+i^{4n-3}z^{4n-2} } = \mathrm {Im} (z^2), \; \forall n \in \mathbb N^*, x \in \mathbb R^*- \{ \pm \sqrt 2/2 \}$$
I have figured out that $i^{4n-3} = i$ for all $n$ in the natural numbers (except 0). So by factoring out $i$ from the denominator and doing some algebraic tricks we finally get to $K(z) = 2x^2 = \mathrm {Im} (z^2)$.
I was wondering if there is another INTERESTING way of solving this
What I have thought so far:
- Geometric series: Let $z^2 = u$ and $a,b: \mathbb N^* \to \mathbb C$ such that $$a_n = u^{2n}, a_1 = u^2, \lambda _1 = u^2 $$ and $$b_k = u^{2k}, b_1 = u, \lambda _2 = u^2$$ So $K(z)$ becomes $$K(u) = \frac {\displaystyle \sum^n_{j=1} a_j }{i \displaystyle \sum^{2n-1}_{m=1} b_m} = \frac {a_1}{ib_1} \frac {\lambda_1^n - 1 } {\lambda_1^{2n-1} - 1} = \frac u i \frac {u^{2n} - 1}{u^{2n-1} - 1}$$
Question: Can I get further from this?
- Analytical geometry: The points $M(z)$ and $P(z^2)$ are moving on the line $\varepsilon : y = x $ and the parabola $C: y = 2x^2$ respectively.
Question: Can I use analytical geometry to solve it?
Finally, every answer is welcome!
An idea using more or less your number (1) - in fact, simply geometric series - :
$$z=x+xi=x(1+i)\implies$$
$$\frac{\sum_{k=1}^n z^{4k}}{\sum_{k=1}^n i^{4k-3}z^{4k-2}}=\frac{\color{red}{\frac{z^4-z^{4(n+1)}}{1-z^4}}}{\frac1{i^3z^2}\color{red}{\frac{i^4z^4-i^{4(n+1)}z^{4(n+1)}}{1-i^4z^4}}}=-iz^2=-ix^2\cdot2i=2x^2$$
since $\;i^{4m}=1\;,\;\;\forall\,m\in\Bbb Z\;$