z+i = 2(1-z')

Let z = a+bi

==> z'= a-bi

Let us substitute:

==> (a+bi) +i= 2(1- (a-bi)

==> a+bi +i= 2(1-a + bi)

==> a+(b+1)i = 2-2a + 2bi

==> a + (b+1)i - 2 + 2a - 2bi= 0

==> 3a + (1-b)i - 2 = 0

==> (3a-2) + (1-b)i = 0

==> 3a-2 = 0 ==> a = 2/3

==> 1-b = 0 ==> b = 1

==> **z = 2/3 + i**

If z is a complex number, we'll write it in the algebraic form:

z = a+b*i

The conjugate of z is z' = a - b*i

We'll re-write the equation:

a + b*i + i = 2(1 - a + b*i)

We'll remove the brackets:

a + i*(b+1) = 2 - 2a + 2b*i

We'll move all terms to one side:

a - 2 + 2a + i*(b+1) - 2b*i = 0

We'll combine the real parts and the imaginary parts:

3a - 2 + i*(b+1-2b) = 0

3a - 2 + i*(1-b) = 0

We could write the right side as:

0 = 0 + 0*i

3a - 2 + i*(1-b) = 0 + 0*i

Because it is an equality, the real part from the left side is equal to the real part from the right side. The same with the imaginary part.

3a - 2 = 0

and

1 - b = 0

3a = 2

**a = 2/3**

**b = 1**

**The complex number is z = (2/3) + i**

z+i = 2(1-z')

Solution:

Let z = a+bi . Then z' = a-bi

Substite in the given equation:

a+bi +i = 2(1- (a-bi))

a+(b+1)i = 2-2a +2bi

Equating real parts a = 2-2a, a+2a = 2 . So a = 2/3.

Equating imaginary parts, b+1 = 2. So b =2-1 =1.

z = a+bi = (2/3)+i.