Question

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

Answer 1

`{(z=-(2i)/(2+i)),(z=-(2i)/(2-i)),(z=-(2i)/3),(z=-2i):}`

Explanation:

`(2z+2i)^4-z^4=((2z+2i)^2+z^2)((2z+2i)^2-z^2)`

or

`(2z+2i+iz)(2z+2i-iz)(2z+2i+z)(2z+2i-z)=0`

so we have:

`{(2z+2i+iz=0),(2z+2i-iz=0),(2z+2i+z=0),(2z+2i-z=0):}`

or

`{(z=-(2i)/(2+i)),(z=-(2i)/(2-i)),(z=-(2i)/3),(z=-2i):}`

Answer 2

`z=-2i`
`z=-2/3i`
`z=2/5-4/5i`
`z=-2/5-4/5i`

Explanation:

We have:

`(2z+2i)^4 = z^4`
`:. ((2z+2i)^2)^2 - (z^2)^2 = 0`

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

`A^2-B^2-=(A+B)(A-B)`

to give:

`((2z+2i)^2 - z^2) ((2z+2i)^2 + z^2) = 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:

`((2z+2i)^2 - z^2) ((2z+2i)^2 - (iz)^2) = 0`

`:. ((2z+2i) - z) ((2z+2i) + z) ((2z+2i) - iz)((2z+2i) + iz) = 0`
`:. (z+2i) (3z+2i) (2z- iz+2i )(2z+ iz+2i ) = 0`

And so we have four solution:

(a) `z+2i = 0 => z=-2i`

(b) `3z+2i = 0 => z=-2/3i`

(c) `2z- iz+2i = 0 => z(2-i)=-2i`

`:. z = (-2i)/(2-i) * (2+i)/(2+i) = (-4i-2i^2)/(4-i^2)=2/5-4/5i`

(d) `2z+ iz+2i = 0 => z(2+i)=-2i`

`:. z = (-2i)/(2+i) * (2-i)/(2-i) = (-4+2i^2)/(4-i^2)=-2/5-4/5i`

Which we can plot on the Argand diagram: