1. ## Complex problem Z=z^2

For what values a and b is the complex number z = a + bi a solution to z=z^2 ?

I tried this in many different ways and il post one of my solutions but i dont know if its correct.
(a+bi) = (a+bi)^2
dividing by a+bi on both sides.
1 = a + bi
This tells me that Z is real and that a = 1 and b = 0.
Is this correct?

2. No, if $\displaystyle z = z^2$

$\displaystyle a + b\,i = (a + b\,i)^2$

$\displaystyle a + b\,i = a^2 + 2ab\,i + b\,i^2$

$\displaystyle a + b\,i = a^2 - b^2 + 2ab\,i$

$\displaystyle 0 + 0i = a^2 - a - b^2 + (2ab - b)i$.

Therefore $\displaystyle a^2 - a - b^2 = 0$ and $\displaystyle 2ab - b = 0$.

Solve these equations simultaneously.

3. Ok.
I recive that a=0.5 and b = +-0.25i
Is this correct?

4. $\displaystyle a$ and $\displaystyle b$ are both real numbers...

5. Ok.
Can you point out where im doing wrong in my calculations?
2ab=b
This leads me to that a = 1/2
a^2 - a -b^2 = 0
Here i add that a=0.5 and recive
-0.25 = b^2
hence i receive that b = +-0.5i

6. Originally Posted by Zamzen
Ok.
Can you point out where im doing wrong in my calculations?
2ab=b
This leads me to that a = 1/2
Strictly that leads to either b= 0 or a= 1/2.

a^2 - a -b^2 = 0

Here i add that a=0.5 and recive
-0.25 = b^2
hence i receive that b = +-0.5i
Yes, which, because b must be a real number, is impossible. It follows that a cannot be 1/2 and that we must have b= 0. Now, from $a^2- a- b^2= 0$, with b= 0, we have $a^2- a= a(a-1)= 0$ so either a= 0 or a= 1.

In fact, we could have said, from the start, that if $z\ne 0$, then, dividing both sides of $z^2= z$ by z, we get $z= 1$. That is, $z^2= z$ if and only if z= 0 or z= 1, real numbers.