We are given the equation $|z| = -1 + \sqrt{6}$, and we are tasked with finding the modulus of $z$ that satisfies the conditions:

• $|z + 2i| = 3$
• $\arg(z) = \frac{\pi}{6}$

Let’s go through the process step by step, utilizing both the cosine rule and polar coordinates.

Step 1: Represent $z$ in Cartesian form

Since we know the argument of $z$ is $\frac{\pi}{6}$, we can write $z$ as: $z = x + iy$ where $\arg(z) = \frac{\pi}{6}$, implying the angle between the vector $z$ and the real axis is $\frac{\pi}{6}$.

Step 2: Use the modulus condition $|z + 2i| = 3$

We know the distance from $z$ to the point $-2i$ on the imaginary axis is 3 units, which leads to the equation: $|z + 2i| = 3$ Substitute $z = x + iy$ into this: $|(x + iy) + 2i| = 3$ $|x + i(y + 2)| = 3$ The modulus of a complex number $a + ib$ is $\sqrt{a^2 + b^2}$, so: $\sqrt{x^2 + (y + 2)^2} = 3$ Square both sides: $x^2 + (y + 2)^2 = 9$ Expand the terms: $x^2 + y^2 + 4y + 4 = 9$ Simplify: x^2 + y^2 + 4y - 5 = 0 \tag{1}

Step 3: Use the argument condition $\arg(z) = \frac{\pi}{6}$

The argument of $z = x + iy$ is given by: $\arg(z) = \tan^{-1} \left( \frac{y}{x} \right)$ Since $\arg(z) = \frac{\pi}{6}$, we have: $\frac{y}{x} = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ Thus: y = \frac{x}{\sqrt{3}} \tag{2}

Step 4: Substitute $y = \frac{x}{\sqrt{3}}$ into equation (1)

Substitute equation (2) into equation (1): $x^2 + \left( \frac{x}{\sqrt{3}} \right)^2 + 4\left(\frac{x}{\sqrt{3}}\right) - 5 = 0$ Simplify each term: $x^2 + \frac{x^2}{3} + \frac{4x}{\sqrt{3}} - 5 = 0$ Multiply the entire equation by 3 to eliminate the fraction: $3x^2 + x^2 + 4\sqrt{3}x - 15 = 0$ Combine like terms: 4x^2 + 4\sqrt{3}x - 15 = 0 \tag{3}

Step 5: Solve the quadratic equation

Now, solve the quadratic equation: $4x^2 + 4\sqrt{3}x - 15 = 0$ Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = 4\sqrt{3}$, and $c = -15$. First, calculate the discriminant: $b^2 - 4ac = (4\sqrt{3})^2 - 4(4)(-15) = 48 + 240 = 288$ So: $x = \frac{-4\sqrt{3} \pm \sqrt{288}}{8} = \frac{-4\sqrt{3} \pm 12\sqrt{2}}{8}$ Simplify: [ x = \frac{-\sqrt{3}}{2} \pm \frac{3\sqrt{6}}