# How to solve this equation where z is a complex number (2z+2i)^4=z^4?

Mar 18, 2017

$\left\{\begin{matrix}z = - \frac{2 i}{2 + i} \\ z = - \frac{2 i}{2 - i} \\ z = - \frac{2 i}{3} \\ z = - 2 i\end{matrix}\right.$

#### Explanation:

${\left(2 z + 2 i\right)}^{4} - {z}^{4} = \left({\left(2 z + 2 i\right)}^{2} + {z}^{2}\right) \left({\left(2 z + 2 i\right)}^{2} - {z}^{2}\right)$

or

$\left(2 z + 2 i + i z\right) \left(2 z + 2 i - i z\right) \left(2 z + 2 i + z\right) \left(2 z + 2 i - z\right) = 0$

so we have:

$\left\{\begin{matrix}2 z + 2 i + i z = 0 \\ 2 z + 2 i - i z = 0 \\ 2 z + 2 i + z = 0 \\ 2 z + 2 i - z = 0\end{matrix}\right.$

or

$\left\{\begin{matrix}z = - \frac{2 i}{2 + i} \\ z = - \frac{2 i}{2 - i} \\ z = - \frac{2 i}{3} \\ z = - 2 i\end{matrix}\right.$

Mar 18, 2017

$z = - 2 i$
$z = - \frac{2}{3} i$
$z = \frac{2}{5} - \frac{4}{5} i$
$z = - \frac{2}{5} - \frac{4}{5} i$

#### Explanation:

We have:

${\left(2 z + 2 i\right)}^{4} = {z}^{4}$
$\therefore {\left({\left(2 z + 2 i\right)}^{2}\right)}^{2} - {\left({z}^{2}\right)}^{2} = 0$

Which is the difference of two squares; and so we use:

${A}^{2} - {B}^{2} \equiv \left(A + B\right) \left(A - B\right)$

to give:

$\left({\left(2 z + 2 i\right)}^{2} - {z}^{2}\right) \left({\left(2 z + 2 i\right)}^{2} + {z}^{2}\right) = 0$

The first factor is again the difference of two square and using ${i}^{2} = - 1$, we can transform the second factor into the same:

$\left({\left(2 z + 2 i\right)}^{2} - {z}^{2}\right) \left({\left(2 z + 2 i\right)}^{2} - {\left(i z\right)}^{2}\right) = 0$

$\therefore \left(\left(2 z + 2 i\right) - z\right) \left(\left(2 z + 2 i\right) + z\right) \left(\left(2 z + 2 i\right) - i z\right) \left(\left(2 z + 2 i\right) + i z\right) = 0$
$\therefore \left(z + 2 i\right) \left(3 z + 2 i\right) \left(2 z - i z + 2 i\right) \left(2 z + i z + 2 i\right) = 0$

And so we have four solution:

(a) $z + 2 i = 0 \implies z = - 2 i$

(b) $3 z + 2 i = 0 \implies z = - \frac{2}{3} i$

(c) $2 z - i z + 2 i = 0 \implies z \left(2 - i\right) = - 2 i$

$\therefore z = \frac{- 2 i}{2 - i} \cdot \frac{2 + i}{2 + i} = \frac{- 4 i - 2 {i}^{2}}{4 - {i}^{2}} = \frac{2}{5} - \frac{4}{5} i$

(d) $2 z + i z + 2 i = 0 \implies z \left(2 + i\right) = - 2 i$

$\therefore z = \frac{- 2 i}{2 + i} \cdot \frac{2 - i}{2 - i} = \frac{- 4 + 2 {i}^{2}}{4 - {i}^{2}} = - \frac{2}{5} - \frac{4}{5} i$

Which we can plot on the Argand diagram: