Question

Find the cube root of $27\left( {\operatorname{Cos} {{30}^o} + i\operatorname{Sin} {{30}^o}} \right)$ that, when represented graphically lies in second quadrant.
A. $3\left( {\operatorname{Cos} {{10}^o} + i\operatorname{Sin} {{10}^o}} \right)$
B. $3\left( {\operatorname{Cos} {{170}^o} + i\operatorname{Sin} {{170}^o}} \right)$
C. $3\left( {\operatorname{Cos} {{100}^o} + i\operatorname{Sin} {{100}^o}} \right)$
D. $3\left( {\operatorname{Cos} {{130}^o} + i\operatorname{Sin} {{130}^o}} \right)$
E. $3\left( {\operatorname{Cos} {{150}^o} + i\operatorname{Sin} {{150}^o}} \right)$

Answer 1

Hint: de Moivre’s theorem is used to calculate the powers of the complex numbers.
As ${\left( {\operatorname{Cos} \theta + i\operatorname{Sin} \theta } \right)^n} = \operatorname{Cos} n\theta + i\operatorname{Sin} n\theta$.

Complete step by step answer:
 The given complex number is,
$y = 27\left( {\operatorname{Cos} {{30}^o} + i\operatorname{Sin} {{30}^o}} \right)$
Taking the cube root of both the sides,
\[
\Rightarrow{y^{\dfrac{1}{3}}} = {27^{\dfrac{1}{3}}}{\left( {\operatorname{Cos} {{30}^o} + i\operatorname{Sin} {{30}^o}} \right)^{\dfrac{1}{3}}} \\
\Rightarrow{y^{\dfrac{1}{3}}} = 3\left( {\operatorname{Cos} \dfrac{{{{30}^o} + {{360}^o}k}}{3} + i\operatorname{Sin} \dfrac{{{{30}^o} + {{360}^o}k}}{3}} \right)......(1) \\
 \]
Put $k = 0$ in equation (1),
$
\Rightarrow{y^{\dfrac{1}{3}}} = 3\left( {\operatorname{Cos} \dfrac{{{{30}^o} + {{360}^o}\left( 0 \right)}}{3} + i\operatorname{Sin} \dfrac{{{{30}^o} + {{360}^o}\left( 0 \right)}}{3}} \right) \\
  {y^{\dfrac{1}{3}}} = 3\left( {\operatorname{Cos} {{10}^o} + i\operatorname{Sin} {{10}^o}} \right) \\
 $
Put $k = 1$ in equation(1),
$
\Rightarrow{y^{\dfrac{1}{3}}} = 3\left( {\operatorname{Cos} \dfrac{{{{30}^o} + {{360}^o}\left( 1 \right)}}{3} + i\operatorname{Sin} \dfrac{{{{30}^o} + {{360}^o}\left( 1 \right)}}{3}} \right) \\
\Rightarrow{y^{\dfrac{1}{3}}} = 3\left( {\operatorname{Cos} {{130}^o} + i\operatorname{Sin} {{130}^o}} \right) \\
 $
Put $k = 2$ in equation(1),
$
\Rightarrow{y^{\dfrac{1}{3}}} = 3\left( {\operatorname{Cos} \dfrac{{{{30}^o} + {{360}^o}\left( 2 \right)}}{3} + i\operatorname{Sin} \dfrac{{{{30}^o} + {{360}^o}\left( 2 \right)}}{3}} \right) \\
\Rightarrow{y^{\dfrac{1}{3}}} = 3\left( {\operatorname{Cos} {{250}^o} + i\operatorname{Sin} {{250}^o}} \right) \\
 $
Hence, the cube root of $27\left( {\operatorname{Cos} {{30}^o} + i\operatorname{Sin} {{30}^o}} \right)$ that lies in second quadrant is $3\left( {\operatorname{Cos} {{130}^o} + i\operatorname{Sin} {{130}^o}} \right)$

Thus, the correct option is (D).

Note: A complex number is a number that can be expressed in the form of $x + iy$ . where, $x$ are real part , represented on X-axis and $iy$ is the imaginary part , represented on Y-axis on the Argand’s plane. The value of ${i^2} = - 1$ .
 A complex number can also be expressed as $e^{ix}$.
Therefore, ${\left( {{e^{ix}}} \right)^n} = {e^{inx}}$
Where $n$ can be any positive, negative integer or it can be a rational number. The term with ‘i’ represents an imaginary part.