z+2z' = 2-i

To find z.

Solution:

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

Therefore the given equation becomes:

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

a+bi+2a-2bi = 2-i

3a -bi = 2-i.

Equate real parts on both sides and similarly imaginary parts:

3a =2 and -bi = -i

3a =2 gives a = 2/3.

-bi = -i = -1*i gives -b = -1. Or b= 1.

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

To determine the complex number z, we'll write first it in the algebraic form:

z = a+b*i

Now we'll write the conjugate z':

z' = a - b*i

We'll re-write the equation, substituting z and z':

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

In the right side we have a complex number, too. We could write the complex number 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**