Q.If $\left(\dfrac{1+i}{1-i}\right)^3-\left(\dfrac{1-i}{1+i}\right)^3=x+iy$, then find $(x, y)$.

The key idea here is simplifying complex fractions and evaluating powers of $i$.

First, simplify the terms $\frac{1+i}{1-i}$ and $\frac{1-i}{1+i}$:

Similarly,

Now, substitute these simplified forms into the given expression:

Evaluate the powers of $i$:

Substitute these back into the expression:

We are given that this expression equals $x+iy$.

Comparing the real and imaginary parts, we find $x=0$ and $y=-2$.

When dealing with complex number expressions involving fractions like $\frac{a+bi}{c+di}$, the most effective first step is almost always to simplify these fractions by rationalizing the denominator. This means multiplying both the numerator and the denominator by the conjugate of the denominator. This process transforms the denominator into a real number, making the entire fraction much easier to work with.

In this problem, we have two such fractions raised to a power. Simplifying them before raising them to the power of $3$ will drastically reduce the complexity of the calculations.

Here's how we approach it:

1. Simplify the first fraction, $\dfrac{1+i}{1-i}$: To simplify this fraction, we multiply the numerator and the denominator by the conjugate of the denominator, which is $1+i$.

Now, we expand the numerator and the denominator. Recall that $(a+b)^2 = a^2+2ab+b^2$ and $(a-b)(a+b) = a^2-b^2$. Also, remember that $i^2 = -1$.

So, the first fraction simplifies to $i$.

> [!TIP]
> The fractions $\frac{1+i}{1-i}$ and $\frac{1-i}{1+i}$ are very common in complex number problems. It's useful to remember their simplified forms: $\frac{1+i}{1-i} = i$ and $\frac{1-i}{1+i} = -i$.

2. Simplify the second fraction, $\dfrac{1-i}{1+i}$:

This fraction is the reciprocal of the first one. We can either calculate it directly or use the result from Step 1.

Using the direct calculation method, we multiply the numerator and denominator by the conjugate of the denominator, which is $1-i$:

Expanding the numerator and denominator:

So, the second fraction simplifies to $-i$.

3. Substitute the simplified forms into the original expression:

The original expression is $\left(\dfrac{1+i}{1-i}\right)^3-\left(\dfrac{1-i}{1+i}\right)^3$.

Substituting our simplified values, this becomes:

4. Calculate the powers of $i$:

   We need to evaluate $i^3$ and $(-i)^3$.

   The powers of $i$ cycle with a period of 4:

   $i^1 = i$

   $i^2 = -1$

   $i^3 = i^2 \cdot i = -1 \cdot i = -i$

   $i^4 = i^2 \cdot i^2 = (-1)(-1) = 1$

   Using this, we find:

   • $i^3 = -i$
   • $(-i)^3 = (-1)^3 \cdot i^3 = -1 \cdot (-i) = i$

5. Perform the subtraction:

   Now substitute these values back into the expression from Step 3:

6. Equate to $x+iy$ and find $(x, y)$: We are given that the expression equals $x+iy$. So, $x+iy = -2i$. To match the form $x+iy$, we can write $-2i$ as $0 + (-2)i$. Comparing the real and imaginary parts: $x = 0$ $y = -2$ Therefore, $(x, y) = (0, -2)$.