how to solve: $(2i)^9z^3=(1+i)^{17}$
we need to find the solutions for the equation. what I did that I simplified the expressions on each side by using the qualities of i to the power n. and then equalizing the imaginary elements on both sides and also equalizing the real ones, and I got that $a=+,-1$ and $b=0$. but I feel its wrong results. someone can check that?
On
The main observation is that $(1+i)^2 = 2i$. So your equation is really just
$$ (1+i)^{18}z^3 = (1+i)^{17}$$
On
I suggest converting the complex vectors to polar form, since multiplication and exponentiation are much easier in polar form.
$$0 + 2i = 2 ~\angle~\pi/2$$ $$1 + i = \sqrt 2 ~\angle~ \pi/4$$
Then just use the results that $(a ~\angle~ b)\times (c~\angle d) = (ac) ~\angle~ (b+d)$, and that $(a ~\angle b)^n = a^n \angle bn$.
$$\begin{align} (2i)^9z^3 &= (1+i)^{17} \\ (2~\angle~ \pi /2)^9~z^3 &= (\sqrt 2 ~\angle~ \pi/4)^{17} \\ (2^9~\angle~ 9\pi /2)~z^3 &= (\sqrt 2^{17} ~\angle~ 17\pi/4) \\ z^3 &= (\sqrt 2^{17} ~\angle~ 17\pi/4)\div (2^9~\angle~ 9\pi /2) \\ z^3 &= \sqrt 2^{17} / \sqrt{2}^{18} ~\angle~ 17\pi/4 - 9\pi /2 \\ z^3 &= 2^{-1/2} ~\angle~ - \pi / 4 \\ \end{align}$$
Now there are 3 cube roots, whose magnitude will be $\sqrt[3]{2^{-1/2}}$ and whose angles are the solutions to $3a_1 = -\pi/4$, $3a_2 = -\pi/4 + 2\pi$ and $3a_3 = -\pi/4 + 4\pi$, giving solutions
$$\begin{cases} z = 2^{-1/6} ~\angle~ -\pi/12 \\ z = 2^{-1/6} ~\angle~ 7\pi/12 \\ z = 2^{-1/6} ~\angle~ 5\pi/4 \end{cases}$$
use that $(2i)^9=512i$ and $(1+i)^{17}=256+256i$ therefore we have $z^3=\frac{1}{2}\frac{1+i}{i}=\frac{1}{2}(1-i)$ $\frac{1+i}{i}=\frac{i-1}{-1}=1-i$