Solve the equation z+z'=z'*z^2+i*z*z' where z is a complex number.

We suppose that z' is the conjugate of the complex number
z.

z = a + bi

z'  =a -
bi

We know that z + z' = a+bi +
a-bi

We'll eliminate like terms and we'll
get:

z + z' = 2a (1)

We'll
calculate the product:

z*z' = (a+bi)(a-bi) = a^2 -
(bi)^2

z*z' = a^2 + b^2
(2)

We'll factorize by z*z' to the right
side:

z+z'= z*z'(z + i)

We'll
substitute the sum and the product by (1) and (2).

2a =
(a^2 + b^2)*(z + i)

We'll divide by (a^2 +
b^2):

2a / (a^2 + b^2) = z +
i

We'll substitute z by a + bi and we'll calculate
z+i.

z+i = a + bi + i

We'll
factorize by i:

z + i = a +
i(b+1)

The equation will
become:

2a / (a^2 + b^2) = a +
i(b+1)

Comparing, we'll
get:

The real parts from both
sides:

2a / (a^2 + b^2) =
a

and

The imaginary
parts:

b+1 = 0

We'll subtract
1:

b = -1

We'll determine
a:

2a / (a^2 + b^2) = a

We'll
divide by a both sides:

2 / (a^2 + b^2) =
1

a^2 + b^2 = 2

Since b =
-1

a^2 + 1 = 2

We'll subtract
1:

a^2 = 1

a1 =
1

a2 = -1

The
solutions of the equation are: {1 - i ; -1 - i }.