Question

find all the solutions to this complex equation...

Answer 1

\[(z + 1)^{8} = (z -1)^{8} \]

complex numbers...

Answer 2

@phi the above question is what i had in my test today.

Answer 3

(z+1 + (z-1))(z+1 - (z-1))((z+1)^2 - (z-1)^2)((z+1)^4 - (z-1)^2)((z+1)^4 + (z-1)^4) = 0

Answer 4

I think z=0 would be a solution.

if we think about them as magnitude and angle, in general they have different lengths (magnitudes).

for the 2 numbers to be at the same spot (after raising to the 8th power), they both must have the same length (and angle).

You would have to work that out to see if there are any solutions with a pure imaginary z

Answer 5

Here is my latest thoughts

Start with 2 complex numbers in the form
\[( A e^{i\theta} )^8 = (B e^{i\psi})^8 \]

let z= a+bi . the length of z+1 is sqrt ((a+1)^b + b^2) = sqrt (a^2 +2a +1 +b^2)
similarly the length of z-1 is sqrt (a^2 -2a +1 +b^2)
equating these 2 we have -a=+a so a=0

This tells us that z must be pure imaginary.
Now to find the roots....

Answer 6

now to find the roots?... :) @phi

Answer 7

Yes, I meant find z's that solve the equation.
I just showed Z is pure imaginary: z= ib where b is an unknown number

z+1 can be written as 1+ ib
z-1 as -1 + ib
we can plot these 2 points: