|z| = -1 + \sqrt{6} show me how you get this using the cosine rule as you find mod(z) that satisfies mod(z+2i)=3 and arg(z)=pi/6

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}}

This solution explores how to find the modulus of complex number z that satisfies |z + 2i| = 3 and arg(z) = pi/6. Detailed steps include using the modulus condition, argument of complex numbers, and solving a quadratic equation. Suitable for undergraduate-level mathematics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation