Question

Express in the form a+ib.
(i). ${{\left( 5-i3 \right)}^{3}}$
(ii). $\dfrac{5+i\sqrt{2}}{1-i\sqrt{2}}$

Answer 1

Hint:-The a+ib is the form to represent complex numbers where a is the real part and b is the imaginary part of the number.
In part (i), the formula for expansion of ${{\left( a-b \right)}^{3}}$ would be required to solve the question and that is as follows
${{\left( a+b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)$
In part (ii), the formula for rationalizing is as follows
$\dfrac{x+y}{x-y}=\dfrac{\left( x+y \right)\left( x-y \right)}{{{x}^{2}}-{{y}^{2}}}$

Complete step-by-step solution -
As mentioned in the question, we are asked to evaluate the two parts and bring them in the form of a+ib.
For part (i), we will use the expansion formula for writing ${{\left( a-b \right)}^{3}}$ that has been given in the hint as follows
${{\left( a+b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab(a-b)$
Now, proceeding with the solution, we get

& {{\left( 5-i3 \right)}^{3}}={{5}^{3}}-{{(i3)}^{3}}-3\times 5\times (i3)(5-3i) \\\ & {{\left( 5-i3 \right)}^{3}}=125-{{(i)}^{3}}27-3\times 5\times (i3)(5-3i) \\\ & {{\left( 5-i3 \right)}^{3}}=125-{{(i)}^{3}}27-45i(5-3i) \\\ & {{\left( 5-i3 \right)}^{3}}=125+(i)27-225i-135 \\\ & {{\left( 5-i3 \right)}^{3}}=-10-198i \\\ \end{aligned}$$ Hence, the part (i) can be represented as -10-i198. Now, for part (ii), we can rationalize the expression and then get to the answer. Now, on rationalizing the expression, we get $$\begin{aligned} & \dfrac{5+i\sqrt{2}}{1-i\sqrt{2}}=\dfrac{\left( 5+i\sqrt{2} \right)\left( 1+i\sqrt{2} \right)}{\left( 1-i\sqrt{2} \right)\left( 1+i\sqrt{2} \right)} \\\ & \dfrac{5+i\sqrt{2}}{1-i\sqrt{2}}=\dfrac{\left( 5+i\sqrt{2} \right)\left( 1+i\sqrt{2} \right)}{\left( {{1}^{2}}-{{\left( i\sqrt{2} \right)}^{2}} \right)} \\\ & \dfrac{5+i\sqrt{2}}{1-i\sqrt{2}}=\dfrac{5+i6\sqrt{2}+2{{i}^{2}}}{\left( {{1}^{2}}-\left( 2{{i}^{2}} \right) \right)} \\\ & \dfrac{5+i\sqrt{2}}{1-i\sqrt{2}}=\dfrac{5+i6\sqrt{2}-2}{\left( 1+2 \right)} \\\ & \dfrac{5+i\sqrt{2}}{1-i\sqrt{2}}=\dfrac{3+i6\sqrt{2}}{3} \\\ & \dfrac{5+i\sqrt{2}}{1-i\sqrt{2}}=1+i2\sqrt{2} \\\ \end{aligned}$$ Hence, this is how we can represent the part (ii) in the form as a+ib. Note: -The students can make an error if they don’t know how to represent an expression in terms of a+ib which is given in the hint as follows The a+ib is the form to represent complex numbers where a is the real part and b is the imaginary part of the number.