To solve the complex quadratic equation:

$3z^2 + (4i - 7)z + (1 - 3i) = 0$

we’ll use the quadratic formula:

$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where:

• $a = 3$
• $b = 4i - 7$
• $c = 1 - 3i$

Step 1: Calculate $b^2 - 4ac$

1. Find $b^2$:

   $b^2 = (4i - 7)^2 = (4i)^2 + 2 \cdot (4i) \cdot (-7) + (-7)^2$ $= 16i^2 - 56i + 49 = 16(-1) - 56i + 49 = -16 - 56i + 49 = 33 - 56i$

2. Calculate $4ac$:

   $4ac = 4 \cdot 3 \cdot (1 - 3i) = 12 \cdot (1 - 3i) = 12 - 36i$

3. Compute $b^2 - 4ac$:

   $b^2 - 4ac = (33 - 56i) - (12 - 36i) = 21 - 20i$

Step 2: Find $\sqrt{b^2 - 4ac}$

To find the square root of $21 - 20i$, we convert it to polar form:

1. Magnitude: $|21 - 20i| = \sqrt{21^2 + (-20)^2} = \sqrt{441 + 400} = \sqrt{841} = 29$

2. Argument: $\theta = \tan^{-1}\left(\frac{-20}{21}\right)$

So $21 - 20i$ in polar form is $29 \text{cis}(\theta)$, and its square root is $\sqrt{29} \text{cis}(\theta/2)$.

Step 3: Solve for $z$

With this square root, substitute back into the quadratic formula to solve for $z$.