We are tasked with finding the relationship between $a$ and $b$, where $z = a + ib$, given:

$2 = \left| \frac{z - 3}{z + 3} \right|$

Let’s solve it step by step:

Step 1: Substitute $z = a + ib$

Substitute $z = a + ib$ into $\frac{z - 3}{z + 3}$:

$\frac{z - 3}{z + 3} = \frac{(a + ib) - 3}{(a + ib) + 3} = \frac{(a - 3) + ib}{(a + 3) + ib}.$

Step 2: Magnitude of a complex number

The magnitude of a complex number $x + iy$ is given by $\sqrt{x^2 + y^2}$. For the given equation, the magnitude is:

$\left| \frac{z - 3}{z + 3} \right| = \frac{\sqrt{(a - 3)^2 + b^2}}{\sqrt{(a + 3)^2 + b^2}}.$

We are given that this magnitude equals 2:

$2 = \frac{\sqrt{(a - 3)^2 + b^2}}{\sqrt{(a + 3)^2 + b^2}}.$

Step 3: Square both sides to eliminate the square roots

Square both sides of the equation:

$4 = \frac{(a - 3)^2 + b^2}{(a + 3)^2 + b^2}.$

Multiply through by the denominator:

$4 \big((a + 3)^2 + b^2\big) = (a - 3)^2 + b^2.$

Step 4: Expand and simplify

Expand both sides of the equation:

1. Expand $4 \big((a + 3)^2 + b^2\big)$: $4 \big((a + 3)^2 + b^2\big) = 4 \big(a^2 + 6a + 9 + b^2\big) = 4a^2 + 24a + 36 + 4b^2.$

2. Expand $(a - 3)^2 + b^2$: $(a - 3)^2 + b^2 = a^2 - 6a + 9 + b^2.$

The equation becomes: $4a^2 + 24a + 36 + 4b^2 = a^2 - 6a + 9 + b^2.$

Step 5: Combine like terms

Simplify by bringing all terms to one side:

$4a^2 - a^2 + 24a + 6a + 4b^2 - b^2 + 36 - 9 = 0.$

$3a^2 + 30a + 3b^2 + 27 = 0.$

Step 6: Factor out 3

Factor out 3 from the equation:

$a^2 + 10a + b^2 + 9 = 0.$

Step 7: Interpret the result

The equation $a^2 + 10a + b^2 + 9 = 0$ describes a relationship between $a$ and $b$. This represents a circle in the complex plane:

$(a + 5)^2 + b^2 = 16.$

Thus, the relationship between $a$ and $b$ is that they lie on a circle centered at $(-5, 0)$ with a radius of 4.

Do you want further clarification, or is there anything specific you’d like to explore further?

Related Questions

1. What is the geometric interpretation of $\left| \frac{z - 3}{z + 3} \right| = 2$?
2. How can we rewrite $a^2 + 10a + b^2 + 9 = 0$ in standard circle form?
3. What happens if the magnitude changes (e.g., $\left| \frac{z - 3}{z + 3} \right| = k$ where $k \neq 2$)?
4. How does the center and radius of the circle depend on the constants in the equation?
5. How can this problem be solved geometrically without algebra?

Tip

Always double-check complex number magnitudes when solving equations involving modulus, as squaring can hide negative values!