No, if
.
Therefore and .
Solve these equations simultaneously.
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?
Strictly that leads to either b= 0 or a= 1/2.
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 , with b= 0, we have so either a= 0 or a= 1.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
In fact, we could have said, from the start, that if , then, dividing both sides of by z, we get . That is, if and only if z= 0 or z= 1, real numbers.