Question

Find all z which satisfy (conjugate z)=z^2

Answer 1

i tried this one... any ideas?

Answer 2

My work just disappeared.

I set (a - bi) = to the square of (a + bi).

I had a = a^2 - b^2 and -b = 2ab.

Then, I decided the answers might be 0 + 0i, 1 + 0i, and -1 + 0i but that's just a conjecture.

Answer 3

i didn't check but I got if z = a + bi,
a=-1/2, b=sqrt(3)/2
checking now...

Answer 4

no, doesn't work.

Answer 5

On the a = a^2, I had that but remembered that all the real parts had to be set equal and the -b^2 is real so a = a^2 - b^2. Then the imaginary parts are set to be equal. so =b = 2ab.

Now, I'm thinking that 0 + 1i and 0 - 1i are solutions.

Answer 6

So far, i'm considering these: 0 + 0i, 1 + 0i, -1 + 0i, and 0 + 1i and 0 - 1i as solutions.

Answer 7

yea. looks good.. 4 solutions

Answer 8

I think there's an elegant way to write a general solution but I don't know what it is.