Express the following in the form a+ bi
`(1+i)^(-3)`

Answer

To express \((1+i)^{-3}\) in the form \(a + bi\), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (1+i)^{-3} = \frac{1}{(1+i)^3} \]

### Step 2: Expand \((1+i)^3\)
Using the binomial theorem or the formula for the cube of a binomial, we have:
\[ (1+i)^3 = 1^3 + 3(1^2)(i) + 3(1)(i^2) + i^3 \]
Calculating each term:
- \(1^3 = 1\)
- \(3(1^2)(i) = 3i\)
- \(3(1)(i^2) = 3(-1) = -3\)
- \(i^3 = i^2 \cdot i = -1 \cdot i = -i\)

Combining these:
\[ (1+i)^3 = 1 + 3i - 3 - i = -2 + 2i \]

### Step 3: Substitute back into the expression
Now substituting back into our expression:
\[ (1+i)^{-3} = \frac{1}{-2 + 2i} \]

### Step 4: Rationalize the denominator
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[ \frac{1}{-2 + 2i} \cdot \frac{-2 - 2i}{-2 - 2i} = \frac{-2 - 2i}{(-2)^2 - (2i)^2} \]
Calculating the denominator:
\[ (-2)^2 = 4 \quad \text{and} \quad (2i)^2 = -4 \quad \Rightarrow \quad 4 - (-4) = 4 + 4 = 8 \]
Thus, we have:
\[ \frac{-2 - 2i}{8} = \frac{-2}{8} + \frac{-2i}{8} = -\frac{1}{4} - \frac{1}{4}i \]

### Final Result
So, we express \((1+i)^{-3}\) in the form \(a + bi\):
\[ (1+i)^{-3} = -\frac{1}{4} - \frac{1}{4}i \]