Find the multiplicative inverse of the following complex number: `(2+sqrt(3)i)^2`

Explanation

To find the multiplicative inverse of the complex number \((2 + \sqrt{3}i)^2\), we will follow these steps:

### Step 1: Calculate \((2 + \sqrt{3}i)^2\)

Using the formula for squaring a binomial, \((a + b)^2 = a^2 + 2ab + b^2\):

Let \(a = 2\) and \(b = \sqrt{3}i\).

\[ (2 + \sqrt{3}i)^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{3}i + (\sqrt{3}i)^2 \]

Calculating each term:

- \(2^2 = 4\)
- \(2 \cdot 2 \cdot \sqrt{3}i = 4\sqrt{3}i\)
- \((\sqrt{3}i)^2 = 3i^2 = 3(-1) = -3\)

Putting it all together:

\[ (2 + \sqrt{3}i)^2 = 4 + 4\sqrt{3}i - 3 = 1 + 4\sqrt{3}i \]

### Step 2: Find the multiplicative inverse

The multiplicative inverse of a complex number \(z = a + bi\) is given by:

\[ \frac{1}{z} = \frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2} \]

In our case, \(z = 1 + 4\sqrt{3}i\), where \(a = 1\) and \(b = 4\sqrt{3}\).

Calculating \(a^2 + b^2\):

\[ a^2 + b^2 = 1^2 + (4\sqrt{3})^2 = 1 + 48 = 49 \]

Now, we can write the multiplicative inverse:

\[ \frac{1}{1 + 4\sqrt{3}i} = \frac{1 - 4\sqrt{3}i}{49} \]

### Step 3: Write in standard form

This can be expressed in the form \(a + bi\):

\[ \frac{1}{1 + 4\sqrt{3}i} = \frac{1}{49} - \frac{4\sqrt{3}}{49}i \]

### Final Answer

The multiplicative inverse of \((2 + \sqrt{3}i)^2\) is:

\[ \frac{1}{49} - \frac{4\sqrt{3}}{49}i \]

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