Solving the equation $(z-2)^{4}+(z+1)^{4}=0$

Question

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

I tried starting by solving

$z^{4}=1$

with the solutions being ,

$1cis (\frac{n\pi }{2})$, where $n = -1, 0, 1, 2$

I am unsure about how to proceed from here, I tried to manipulate the original equation so it would fit with the solution of $z^{4}=1$ somehow, but it hasn't worked.

Can someone guide me in the correct direction? with preferably the same approach as I have taken?

Answer 1

We write the given equation on this form

$$\left(\frac{z-2}{z+1}\right)^4=-1=e^{i\pi}$$ so $$\frac{z-2}{z+1}=\exp\left(\frac{i\pi+2ik\pi}{4}\right),\qquad k=0,1,2,3$$ finally solving the last equation for the unknown $z$ is a piece of cake.

Answer 2

Another way :

Set $\displaystyle2y=z-2+z+1=2z-1\implies z-2=\frac{2y-3}2$ and $\displaystyle z+1=\frac{2y+3}2$

$\displaystyle\implies(2y-3)^4+(2y+3)^4=0\iff(2y)^4+6(2y)^2(3)^2+(3)^4=0$

Answer 3

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

Difference of squares

$$\left((z-2)^2\right)^2 - \left(i(z+1)^2\right)^2=0$$ $$ \left((z-2)^2 - i(z+1)^2\right) \left((z-2)^2 + i(z+1)^2\right) =0$$ ... and again... $$ \left((z-2)^2 - \left(\sqrt{i}(z+1)\right)^2\right) \left((z-2)^2 - \left(i\sqrt{i}(z+1)\right)^2\right) =0$$ $$ \left(z-2 - \sqrt{i}(z+1)\right) \left(z-2 + \sqrt{i}(z+1)\right) \left(z-2 - i\sqrt{i}(z+1)\right) \left(z-2 + i\sqrt{i}(z+1)\right) =0$$ stuff $$\begin{cases} z-2 - \sqrt{i}(z+1) = 0 \\ z-2 + \sqrt{i}(z+1) = 0 \\ z-2 - i\sqrt{i}(z+1) = 0 \\ z-2 + i\sqrt{i}(z+1) = 0 \\ \end{cases}$$ $$\begin{cases} (1 - \sqrt{i})z = 2 + \sqrt{i} \\ (1 + \sqrt{i})z = 2 - \sqrt{i} \\ (1 - i \sqrt i )z = 2 + i\sqrt{i} \\ (1 + i \sqrt i )z = 2 - i\sqrt{i} \\ \end{cases}$$ $$\begin{cases} z = \frac{2 + \sqrt{i}}{1 - \sqrt{i}} \\ z = \frac{2 - \sqrt{i}}{1 + \sqrt{i}} \\ z = \frac{2 + i\sqrt{i}}{1 - i \sqrt i } \\ z = \frac{2 - i\sqrt{i}}{1 + i \sqrt i } \\ \end{cases}$$