Solve, in complex numbers, the equation $(z − i)^{3} =\overline{z} + i$

Question

Solve, in complex numbers, the equation $(z − i)^{3} = \overline{z} + i$

I tried raising left side to the power but after doing that I still wasn't able to solve that. Any tips? I heard something about Euler's formula but as far as I remember we have never used it during tutorials

Answer 1

it is $$x^3-3xy^2+6xy-4x+i(3x^2y-3x^2-y^3+3y^2-2y)=0$$ and you must solve the System $$x^3-3xy^2+6xy-4x=0$$ and $$3x^2y-3x^2-y^3+3y^2-2y=0$$ the first equation can be written as $$x(x^2-3y^2+6y-4)=0$$ and the second as $$3x^2(y-1)-y^3+3y^2-2y=0$$ does this help? from the first equation we get $$x=0$$ or $$x^2-3y^2+6y-4=0$$ we can eliminate $$x^2$$ from the second equation $$x^2=\frac{y^3-3y^2+2y}{3(y-1)}$$ if $$y\neq 1$$ doing so we have on equation for $y$ $$-8y^3+24y^2-29y+14=0$$

Answer 2

Hint: taking the conjugates on both sides gives $(\bar z + i)^{3} = z - i$, then substituting $\bar z = (z − i)^{3} - i $ from the original equation:

$$\require{cancel} ((z − i)^{3} - \bcancel{i} + \bcancel{i})^{3} = z - i \quad\iff\quad (z-i)\big((z-i)^8-1\big) = 0 $$

Answer 3

Set $u=z-i$, and note the equation can be re-written as $$u^3=\bar {\rule{0pt}{1.35ex} u}$$ One obvious solution is $u=0$, i.e. $\color{red} {z=i}$.

Now suppose $u\ne 0$, and use the exponential form of $u$. The equation becomes $$r^3\mathrm e^{3i\theta}=r\mathrm e^{-i\theta}\iff r^2\mathrm e^{4i\theta}=1\iff\begin{cases}r=1,\\4\theta\equiv 0\mod 2\pi\end{cases}\iff\begin{cases}r=1,\\\theta\equiv 0\mod \dfrac\pi2.\end{cases}$$ So $u\in\bigl\{1,-1,i,-i\bigr\}$, and $$z\in\bigl\{\color{red}{1+i},\color{red}{-1+i},\color{red}{2i},\color{red}{0}\bigr\}.$$