If `((1+i)/(1-i))^3-((1-i)/(1+i))^3 =a+ib` find a and b

Options

A. 0 and 2
B. 0 and -2
C. 2 and 0
D. 2 and 2

Answer

To solve the problem \(\left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3 = a + ib\), we will follow these steps:

### Step 1: Simplify the fractions
We start by simplifying \(\frac{1+i}{1-i}\) and \(\frac{1-i}{1+i}\). We can multiply the numerator and denominator of each fraction by the conjugate of the denominator.

For \(\frac{1+i}{1-i}\):
\[ \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i \]

For \(\frac{1-i}{1+i}\):
\[ \frac{1-i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i \]

### Step 2: Substitute back into the equation
Now we substitute back into the original equation:
\[ \left(i\right)^3 - \left(-i\right)^3 \]

### Step 3: Calculate the cubes
Calculating the cubes:
\[ i^3 = -i \quad \text{and} \quad (-i)^3 = -(-i) = i \]
Thus, we have:
\[ -i - i = -2i \]

### Step 4: Write in the form \(a + ib\)
Now we can express \(-2i\) in the form \(a + ib\):
\[ 0 - 2i = 0 + (-2)i \]
This gives us \(a = 0\) and \(b = -2\).

### Final Answer
Thus, the values of \(a\) and \(b\) are:
\[ a = 0, \quad b = -2 \]