Determine the complex number z if (3z-2z')/6=-5? z' is the conjugate of z.

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We'll write the rectangular form of the complex number z:

z = a + bi

Since z' is the conjugate of z, then z' = a - b.

To determine z, we'll have to find out it's coefficients:

We'll multiply by 6 both sides:

3(a + bi) - 2(a - bi) = - 30

We'll remove the brackets:

3a + 3bi - 2a + 2bi = - 30

We'll move all terms to the left side:

3a + 3bi - 2a + 2bi + 30 = 0

We'll combine real parts and imaginary parts:

a + 5bi + 30 = 0

The real part of the complex number from the left side is:

Re(z) = a + 30

The real part of the complex number from the right side is:

Re(z) = 0

Comparing, we'll get:

a + 30 = 0

a = -30

The imaginary part of the complex number from the left side is:

Im(z) = 5b

The imaginary part of the complex number from the right side is:

Im(z) = 5b

Comparing, we'll get:

5b = 0

b = 0

**The requested complex number z is: ****z = -30 (whose imaginary part is 0).**

We have to find the complex number z given that (3z - 2z')/6 = -5

Let z = a + ib, z' = a - ib

(3z - 2z')/6 = -5

=>(3(a + ib) - 2(a - ib)) = -30

=> 3a + 3ib - 2a + 2ib = -30

=> a + 5ib = -30

We see that for the imaginary part 5ib there is no equivalent on the right hand side, so b = 0

a = -30

**The number z = -30**

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