Question

If \[a\] and \[b\] are real numbers such that \[{\left( {2 + \alpha } \right)^4} = a + b\alpha \] , where \[\alpha = \dfrac{{\left( { - 1 + i\sqrt 3 } \right)}}{2}\] . Then what is the value of \[\left( {a + b} \right)\]?
A. \[33\]
B. \[57\]
C. 9
D. \[24\]

Answer 1

Hint: First, substitute the value of \[\alpha \] in the given equation and simplify it. Then solve the left-hand side of the equation by converting the complex number into the polar form. After that, convert the polar form into the exponential form of a complex number and simplify it. Again, convert the terms on the left-hand side into the polar form. Substitute the values of the trigonometric angles and solve the equation. In the end, equate the terms of both sides to get the required answer.

Formula used:
The polar form of a complex number \[z = a + ib\] is, \[z = r\left( {cos\theta + isin\theta } \right)\] . where \[r = \sqrt {{a^2} + {b^2}} \].
The exponential form of a complex number \[z = a + ib\] is, \[z = r{e^{i\theta }}\].

Complete step by step solution:
The given equation is \[{\left( {2 + \alpha } \right)^4} = a + b\alpha \], where \[a,b \in R\] and \[\alpha = \dfrac{{\left( { - 1 + i\sqrt 3 } \right)}}{2}\].
Let’s substitute the value of \[\alpha \] in the given equation.
\[{\left( {2 + \dfrac{{\left( { - 1 + i\sqrt 3 } \right)}}{2}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Simplify the left-hand side of the above equation.
\[{\left( {\dfrac{{4 - 1 + i\sqrt 3 }}{2}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
\[ \Rightarrow {\left( {\dfrac{{3 + i\sqrt 3 }}{2}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
\[ \Rightarrow {\left( {\dfrac{3}{2} + \dfrac{{i\sqrt 3 }}{2}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Factor out the common value from the left-hand side.
\[{\left( {\sqrt 3 } \right)^4}{\left( {\dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
\[ \Rightarrow 9{\left( {\dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Now convert the left-hand side of the above equation in the polar form.
\[9{\left( {\cos\dfrac{\pi }{6} + i\sin\dfrac{\pi }{6}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Convert the polar form into the exponential form of a complex number.
\[9{\left( {{e^{i\dfrac{\pi }{6}}}} \right)^4} = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Apply the exponent property \[{\left( {{a^m}} \right)^n} = {a^{mn}}\] .
\[9\left( {{e^{i\dfrac{{2\pi }}{3}}}} \right) = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Again, convert the above equation into the polar form.
\[9\left( {\cos\dfrac{{2\pi }}{3} + i\sin\dfrac{{2\pi }}{3}} \right) = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Substitute the values of the trigonometric angles in the above equation.
\[9\left( {\dfrac{{ - 1}}{2} + i\dfrac{{\sqrt 3 }}{2}} \right) = a + b\left( {\dfrac{{ - 1 + i\sqrt 3 }}{2}} \right)\]
Now simplify the above equation.
\[\dfrac{{ - 9}}{2} + i\dfrac{{9\sqrt 3 }}{2} = a - \dfrac{b}{2} + \dfrac{{ib\sqrt 3 }}{2}\]
Now equate the real part and imaginary part of the above equation.
We get,
\[\dfrac{{ - 9}}{2} = a - \dfrac{b}{2}\] \[.....\left( 1 \right)\]
\[i\dfrac{{9\sqrt 3 }}{2} = \dfrac{{ib\sqrt 3 }}{2}\] \[.....\left( 2 \right)\]
From the equation \[\left( 2 \right)\], we get
\[b = 9\]
Now substitute the value of \[b\] in the equation \[\left( 1 \right)\]
\[\dfrac{{ - 9}}{2} = a - \dfrac{9}{2}\]
\[ \Rightarrow a = \dfrac{9}{2} - \dfrac{9}{2}\]
\[ \Rightarrow a = 0\]
Therefore,
\[a + b = 0 + 9\]
\[ \Rightarrow a + b = 9\]
Hence the correct option is C.

Note: Students often get confused with the different types of a complex number.
Rectangular form: \[z = x + iy\]
Polar form: \[z = r\left( {cos\theta + isin\theta } \right)\], where \[r = \sqrt {{a^2} + {b^2}} \]
Exponential form: \[z = r{e^{i\theta }}\], where \[{e^{i\theta }} = \cos\theta + i\sin\theta \]