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

==>...

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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**