z + 3z' = 3i

z = a+bi

First ley us determine z':

==> z' = a-bi

Now substitute:

z + 3z' = 3i

(a+bi) + 3(a-bi) = 3i

==> a + bi + 3a - 3bi = 3i

==> 4a - 2bi = 3i

==> 4a = 0 ==> a = 0

==> -2bi = 3i

==> -2b = 3 ==> b = -3/2

==> z= 0 + (-3/2)i

==> Z = -(3/2)i

z+3z' = 3i .

To find z.

Solution:

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

So the given equation becomes:

a+bi+3(a-bi) = 3i = 0+3i

4a -2 bi = 0 +3i. Equate real parts on both sides . Equate imagonary parts on both sides.

4a = 0 and -2bi = 3i

a = 0 b = -3/2.

So z = a+bi = 0+(-3/2)i = -1.5i

z = -1.5i

Considering that z'=a-b*i is the conjugate of z, the given relation could be re-written:

z+3z'=3i

a+b*i +3(a-b*i)=3i

a+b*i + 3a - 3*b*i - 3i = 0

We'll combine real terms and imaginary terms:

4a + i*(b-3b-3) = 0

4a + i*(-2b-3) = 0

We'll consider the right side as a complex number:

0 = 0 + 0*i

4a + i*(-2b-3) = 0 + 0*i

Since the both sides are equivalent, we'll write:

4a = 0

**a = 0**

-2b-3 = 0

-2b = 3

**b = - 3/2**

**The complex number z will be:**

**z = 0 + i*(-3/2)**