What is the complex number conjugate with itself? What is the complex number conjugate with its square?

Let the complex number be a + bi, a and B be real numbers
1, complex number conjugate with itself
a + bi = a - bi,
b = -b,
2b = 0,
b = 0.
Therefore, a complex number conjugate with itself is a real number
2, the complex number conjugate with the square of itself
(a + bi)^2 = a^2 - b^2 + 2abi
a + bi = a^2 - b^2 - 2abi,
In other words,
a = a^2 - b^2,
b = -2ab.
if
b = 0.
Then, a = a ^ 2, a (A-1) = 0,
A = 0 or a = 1
if
B is not equal to 0
Then, 1 = - 2A, a = - 1 / 2
b^2 = a^2 - a = 1/4 + 1/2 = 3/4.
B = 3 ^ (1 / 2) / 2, or B = - 3 ^ (1 / 2) / 2
All in all,
A = 0, B = 0, the complex number conjugate with its square is 0,
Or, a = 1, B = 0, and the complex number conjugate with its square is 1,
Or, a = - 1 / 2, B = 3 ^ (1 / 2) / 2, and the complex number conjugated with its square is [- 1 + I3 ^ (1 / 2)] / 2,
Or, a = - 1 / 2, B = - 3 ^ (1 / 2) / 2, the complex number conjugated with the square of itself is [- 1-i3 ^ (1 / 2)] / 2

Solving a mathematical problem about complex number
If the complex Z satisfies (Z-I) I = 2 + I, then z =?

z-i=1-2i
z=1-i

How to transform linear equation into complex form
Write the linear equation AX + by + C = 0 (a * a + b * b =? 0) in complex form (Note: x + iy = z)

From the hint, we can see that it means to change X and Y into the algebraic expression of Z, and then substitute it into the equation, and then we can get the equation about Z, which is written in complex form. But x + iy = Z, there is only one Z, so to solve x and y, we need to use the conjugate of complex number, which is denoted as [Z], that is, x + iy = ZX iy = [Z] to solve x = (Z

What is the complex number conjugate with itself? What is the complex number conjugate with its square?