Question

Express the given complex number ${\left[ {\dfrac{1}{3} + 3i} \right]^3}$ in the form of $a + ib$.
$\left( 1 \right)$ $\dfrac{{343}}{{34}} + 23i$
$\left( 2 \right)$ $\dfrac{{ - 343}}{{34}} - 23i$
$\left( 3 \right)$ $\dfrac{{242}}{{27}} + 26i$
$\left( 4 \right)$ $\dfrac{{ - 242}}{{27}} - 26i$

Answer 1

We have to express the value of the given complex number in the form of $a + ib$. We solve this question using the concept of values of complex numbers. We should also have the knowledge of the formula of the cube of sum of two numbers. We should also know the value of iota function for different powers. First, we will expand the given expression by using the formula of a cube of sum of two numbers and then by putting the values of powers of iota we will simplify the expression and then we will split the real terms and complex terms separately. And hence on solving the values we will obtain a relation of the given expression in terms of $a + ib$.

Complete step-by-step solution:
Given :
The given expression is ${\left[ {\dfrac{1}{3} + 3i} \right]^3}$
Now, we also know that the formula of cube of sum of two numbers is given as :
${(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$
Using the formula, we get the value o the expression as :
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = {\left( {\dfrac{1}{3}} \right)^3} + {\left( {3i} \right)^3} + 3{\left( {\dfrac{1}{3}} \right)^2}\left( {3i} \right) + 3\left( {\dfrac{1}{3}} \right){\left( {3i} \right)^2}$
On further simplifying, we get
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = \left( {\dfrac{1}{{27}}} \right) + 27{i^3} + 3\left( {\dfrac{1}{9}} \right)\left( {3i} \right) + 3\left( {\dfrac{1}{3}} \right)9{i^2}$
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = \left( {\dfrac{1}{{27}}} \right) + 27{i^3} + i + 9{i^2}$
Now, we also know that the values for different powers of iota is given as :
${i^2} = - 1$
${i^3} = - i$
Putting these values we get the expression as :
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = \left( {\dfrac{1}{{27}}} \right) + 27\left( { - i} \right) + i + 9\left( { - 1} \right)$
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = \dfrac{1}{{27}} - 9 - 27i + i$
Now solving the expression by taking L.C.M., we get
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = \dfrac{{1 - 9 \times 27}}{{27}} - 26i$
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = \dfrac{{1 - 243}}{{27}} - 26i$
Further, we get the value as :
${\left[ {\dfrac{1}{3} + 3i} \right]^3} = \dfrac{{ - 242}}{{27}} - 26i$
The simplified relation is in the form of $a + ib$, where a is the real part of the simplified expression and b is the complex part of the simplified relation.
Thus, we express ${\left[ {\dfrac{1}{3} + 3i} \right]^3}$ in the form of $a + ib$ as $\dfrac{{ - 242}}{{27}} - 26i$.
Hence, the correct option is $\left( 4 \right)$.

Note: A number of the form $a + ib$, where a and b are real numbers, is called a complex number, a is called the real part and b is called the imaginary part of the complex number. Every real number can be represented in terms of complex numbers but the converse is not true. Since ${b^2} - 4ac$ determines whether the quadratic equation a ${x^2} + bx + c = 0$. If ${b^2} - 4ac < 0$ then the equation has imaginary roots.