\( |4 z-6|=|z+3| \) The solution set is

We start with the equation   |4z – 6| = |z + 3|. Step 1. Write z in the form x + yi, where x and y are real numbers. Step 2. Express each modulus in terms of x and y:   |4z – 6| = |4(x + yi) – 6| = |(4x – 6) + 4yi| = √[(4x – 6)² + (4y)²],   |z + 3| = |(x + 3) + yi| = √[(x + 3)² + y²]. Step 3. Square both sides to eliminate the square roots:   (4x – 6)² + 16y² = (x + 3)² + y². Step 4. Expand and simplify:   (16x² – 48x + 36) + 16y² = x² + 6x + 9 + y². Subtract the right side from the left:   16x² – 48x + 36 + 16y² – x² – 6x – 9 – y² = 0. Combine like terms:   15x² – 54x + 27 + 15y² = 0. Step 5. Divide the whole equation by 15 to simplify:   x² + y² – (54/15)x + (27/15) = 0,   x² + y² – (18/5)x + (9/5) = 0. Step 6. Complete the square for the x-terms:   x² – (18/5)x = (x – 9/5)² – (9/5)²  since (x – 9/5)² = x² – (18/5)x + (81/25). Substitute back into the equation:   (x – 9/5)² – (81/25) + y² + (9/5) = 0. Step 7. Bring the constants to the right-hand side:   (x – 9/5)² + y² = (81/25) – (9/5). Convert (9/5) to twenty-fifths:   (9/5) = (45/25), Thus,   (x – 9/5)² + y² = (81/25 – 45/25) = 36/25. Step 8. Recognize the equation of a circle:   (x – 9/5)² + y² = (6/5)². Thus, in terms of the complex number z, the solution set is the circle given by   |z – 9/5| = 6/5. So, the solution set is all complex numbers z such that |z – 9/5| = 6/5.