Simplify (1 + 2 /𝑖 ) (1 + 3 /𝑖 ) (2 + 𝑖) ^−3 and express it in the form of a +ib.

We are tasked with simplifying the expression \[
\left(1 + \frac{2}{i}\right)\left(1 + \frac{3}{i}\right)(2 + i)^{-3}
\]
and expressing the result in the form \( a + ib \), where \(a\) and \(b\) are real numbers.

### Step 1: Simplify the fractions
Using the fact that \( \frac{1}{i} = -i \), we can rewrite the fractions:
\[
1 + \frac{2}{i} = 1 - 2i, \quad 1 + \frac{3}{i} = 1 - 3i.
\]
Thus, the expression becomes:
\[
(1 - 2i)(1 - 3i)(2 + i)^{-3}.
\]

### Step 2: Multiply the first two terms
Next, we multiply \( (1 - 2i)(1 - 3i) \) using the distributive property:
\[
(1 - 2i)(1 - 3i) = 1(1 - 3i) - 2i(1 - 3i).
\]
Expanding the terms:
\[
= 1 - 3i - 2i + 6i^2.
\]
Since \( i^2 = -1 \), we have:
\[
1 - 3i - 2i + 6(-1) = 1 - 5i - 6 = -5 - 5i.
\]

So, the product of the first two terms is \( -5 - 5i \).

### Step 3: Simplify the third term
Now, we need to simplify \( (2 + i)^{-3} \). First, find the modulus and argument of \( 2 + i \).

- The modulus is:
  \[
  |2 + i| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}.
  \]
- The argument \( \theta \) is:
  \[
  \theta = \tan^{-1}\left(\frac{1}{2}\right).
  \]

Using De Moivre’s theorem:
\[
(2 + i)^{-3} = \frac{1}{(2 + i)^3}.
\]
First, we cube \( 2 + i \

Simplify (1 + 2/i) (1 + 3/i) (2 + i)^−3 and express it in the form of a + ib.

This page walks through simplifying the complex expression (1 + 2/i)(1 + 3/i)(2 + i)^−3 and expressing the result in the form a + ib. It covers key concepts such as complex number multiplication, De Moivre's Theorem, and converting complex fractions.

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