Question

Using the given equation find the (x, y)
${{\left( \dfrac{1+i}{1-i} \right)}^{3}}-{{\left( \dfrac{1-i}{1+i} \right)}^{3}}=x+iy$

Answer 1

Hint: Rationalize the terms inside the bracket by multiplying and dividing by 1+i or 1-i.
After rationalizing, solve it like a general algebraic equation.

Complete step-by-step solution -

The given equation is:
${{\left( \dfrac{1+i}{1-i} \right)}^{3}}-{{\left( \dfrac{1-i}{1+i} \right)}^{3}}=x+iy.....\left( 1 \right)$
We aim to find: (x, y)
To solve the left hand side, first we need to rationalize the terms.
Firstly, take the first term on the left hand side.
${{\left( \dfrac{1+i}{1-i} \right)}^{3}}$
To rationalize this term we need to multiply and divide with 1 + i inside the bracket.
By multiplying and dividing with 1 + i inside the bracket, we get:
${{\left( \dfrac{1+i}{1-i}.\dfrac{1+i}{1+i} \right)}^{3}}$
We need use a basic algebraic identity in denominator:
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$
By using the above identity, we get:

& {{\left( \dfrac{{{\left( 1+i \right)}^{2}}}{1-{{i}^{2}}} \right)}^{3}} \\\ & \text{we know that }{{\text{i}}^{2}}=-1 \\\ \end{aligned}$$ By simplifying, we get: $$\begin{aligned} & {{\left( \dfrac{1+{{i}^{2}}+2i}{1-{{i}^{2}}} \right)}^{3}} \\\ & {{\left( \dfrac{1-1+2i}{1-\left( -1 \right)} \right)}^{3}} \\\ \end{aligned}$$ By cancelling the common terms, we get: $${{\left( \dfrac{2i}{2} \right)}^{3}}={{i}^{3}}$$ By simplifying and substituting value of i, we get: $${{\left( \dfrac{1+i}{1-i} \right)}^{3}}=-i.....\left( 2 \right)$$ Next take the second term on the left hand side. $${{\left( \dfrac{1-i}{1+i} \right)}^{3}}$$ To rationalize this term we need to multiply and divide with 1 - i inside the bracket. By multiplying and dividing with 1 - i inside the bracket, we get: $${{\left( \dfrac{1-i}{1+i}.\dfrac{1-i}{1-i} \right)}^{3}}$$ We need use a basic algebraic identity in denominator: $$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$$ By using the above identity, we get: $$\begin{aligned} & {{\left( \dfrac{{{\left( 1-i \right)}^{2}}}{1-{{i}^{2}}} \right)}^{3}} \\\ & \text{we know that }{{\text{i}}^{2}}=-1 \\\ \end{aligned}$$ By simplifying, we get: $$\begin{aligned} & {{\left( \dfrac{1+{{i}^{2}}-2i}{1-{{i}^{2}}} \right)}^{3}} \\\ & {{\left( \dfrac{1-1-2i}{1-\left( -1 \right)} \right)}^{3}} \\\ \end{aligned}$$ By cancelling the common terms, we get: $${{\left( -\dfrac{2i}{2} \right)}^{3}}=-{{i}^{3}}$$ By simplifying and substituting value of i, we get: $${{\left( \dfrac{1-i}{1+i} \right)}^{3}}=i.....\left( 3 \right)$$ By using equation (2), equation (3) we find left hand side as follows: $${{\left( \dfrac{1+i}{1-i} \right)}^{3}}-{{\left( \dfrac{1-i}{1+i} \right)}^{3}}=-i-\left( i \right)$$ So we find the left hand side to be 2i. By equating left hand side to right hand side, we get: 2i = x + iy. We can write 2i as 0 + 2i. So the equation turns out to be: x + iy = 0 + 2i By comparing, we get: x = 0 y = 2 (x, y) = (0, 2) $$\therefore $$The required values of x, y are 0, 2 respectively. Note: When you see a complex number in division format the first thing to do must be rationalizing. This makes the question easy to solve. Take care of sign while subtracting equation (2) and equation (3)