Question

Find the one of the values of \[{\left( {\dfrac{{\left( {1 + i} \right)}}{{\sqrt 2 }}} \right)^{\dfrac{2}{3}}}\].
A. \[\sqrt 3 + i\]
B. \[ - i\]
C. \[i\]
D. \[ - \sqrt 3 + i\]

Answer 1

Hint First, we write given complex number in polar form \[\cos\theta + i\sin\theta \]. After that, convert the polar form into the exponential form \[{e^{i\theta }}\] and simplify it. Then apply the general formula \[{e^{i\theta }}=\cos\left( {2n + 1} \right)\theta + i\sin\left( {2n + 1} \right)\theta \]. After that we will substitute n=1 in the expression. Solve the expression to get the required answer.

Formula used
\[\cos\left( {\dfrac{\pi }{4}} \right) = \sin\left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\]
A complex number \[\cos\theta + i\sin\theta \] can be written as \[{e^{i\theta }}\].
The general form of a complex number \[{e^{i\theta }}\] is \[\cos\left( {2n + 1} \right)\theta + i\sin\left( {2n + 1} \right)\theta \].

Complete step by step solution:
The given complex number is \[{\left( {\dfrac{{\left( {1 + i} \right)}}{{\sqrt 2 }}} \right)^{\dfrac{2}{3}}}\].
Let consider,
\[z = {\left( {\dfrac{{\left( {1 + i} \right)}}{{\sqrt 2 }}} \right)^{\dfrac{2}{3}}}\]
Simplify the equation.
\[z = {\left( {\dfrac{1}{{\sqrt 2 }} + \dfrac{i}{{\sqrt 2 }}} \right)^{\dfrac{2}{3}}}\]
Rewrite the above complex number in the polar form.
\[z = {\left( {\cos\left( {\dfrac{\pi }{4}} \right) + i\sin\left( {\dfrac{\pi }{4}} \right)} \right)^{\dfrac{2}{3}}}\] [ Since \[\cos\left( {\dfrac{\pi }{4}} \right) = \sin\left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\]]
Rewrite the above complex number in the exponential form.
\[z = {\left( {{e^{i\dfrac{\pi }{4}}}} \right)^{\dfrac{2}{3}}}\]
\[ \Rightarrow \]\[z = {e^{i\dfrac{\pi }{6}}}\] [ Since \[{\left( {{a^m}} \right)^n} = {a^{mn}}\]]
Rewrite the complex number in the trigonometry form:
\[ \Rightarrow z = \cos \left( {2n + 1} \right)\dfrac{\pi }{6} + i\sin \left( {2n + 1} \right)\dfrac{\pi }{6}\]
Substitute \[n = 1\] in the above equation:
\[ \Rightarrow z = \cos \left( {3 \cdot \dfrac{\pi }{6}} \right) + i\sin \left( {3 \cdot \dfrac{\pi }{6}} \right)\]
\[ \Rightarrow z = \cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}\]
\[ \Rightarrow z = 0 + i\left( 1 \right)\]
\[ \Rightarrow z = 0 + i\]
\[ \Rightarrow z = i\]
Hence the correct option is C.

Note: Students often do a common mistake to solve this question. They use \[\cos\theta + i\sin\theta ={e^{i\theta }}\] instead of the general form \[{e^{i\theta }}= \cos\left( {2n + 1} \right)\theta + i\sin\left( {2n + 1} \right)\theta \]. So that they get \[\dfrac{\sqrt{3}}{2}+i\dfrac{1}{2}\] which is an incorrect answer.