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Question: The real part of \( {\left( {i - \sqrt 3 } \right)^{13}} \) is : a) \( {2^{ - 10}}\sqrt 3 \) b)...

The real part of (i−3)13{\left( {i - \sqrt 3 } \right)^{13}} is :
a) 2−103{2^{ - 10}}\sqrt 3
b) −2123- {2^{12}}\sqrt 3
c) 2−123{2^{ - 12}}\sqrt 3
d) −2−123- {2^{ - 12}}\sqrt 3
e) −2103- {2^{10}}\sqrt 3

Explanation

Solution

Hint : We will use the cube root of unity to solve the problem. The given expression looks similar to one of the cube roots of unity ( w2{w^2} ) so we will reduce it to that with required algebraic manipulations. After that we will use the properties of the root to find the answer.
Formula used:
The solutions to the equation x3−1=0{x^3} - 1 = 0 are called the cube root of unity.
The cube roots of unity are 1,w,w21,w,{w^2} where, w=−1+i32,w2=−1−i32w = \dfrac{{ - 1 + i\sqrt 3 }}{2},{w^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}

  1. wn=wk{w^n} = {w^k} , where k is the remainder after dividing n with 3.
  2. in=ik{i^n} = {i^k} , where k is the remainder after dividing n with 4 and i=−1i = \sqrt { - 1} .

Complete step-by-step answer :
The given expression is (i−3)13{\left( {i - \sqrt 3 } \right)^{13}}
Let y=i−3y = i - \sqrt 3
Now we will reduce this expression to w2{w^2} which is the cube root of unity.
Let’s multiply ‘I’ and divide ‘2’ in the above equation.
⇒iy=−1−i3\Rightarrow iy = - 1 - i\sqrt 3 [Multiplying ‘i’ throughout the equation]
⇒iy2=−1−i32\Rightarrow \dfrac{{iy}}{2} = \dfrac{{ - 1 - i\sqrt 3 }}{2} [Dividinging 2 throughout the equation]
⇒iy2=w2\Rightarrow \dfrac{{iy}}{2} = {w^2} where w2{w^2} is one of the cube root of unity.
Now we have,
iy2=w2\dfrac{{iy}}{2} = {w^2}
⇒iy=2w2\Rightarrow iy = 2{w^2}
⇒(iy)13=(2w2)13\Rightarrow {\left( {iy} \right)^{13}} = {\left( {2{w^2}} \right)^{13}}
⇒i13×y13=213(w2)13\Rightarrow {i^{13}} \times {y^{13}} = {2^{13}}{\left( {{w^2}} \right)^{13}}
⇒i12×i×y13=213×w16\Rightarrow {i^{12}} \times i \times {y^{13}} = {2^{13}} \times {w^{16}}
⇒1×i×y13=213×w2\Rightarrow 1 \times i \times {y^{13}} = {2^{13}} \times {w^2}
⇒y13=−i(213×w2)\Rightarrow {y^{13}} = - i\left( {{2^{13}} \times {w^2}} \right) [Multiplying ‘-i’ throughout the equation]
As w2=−1−i32{w^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2} so we can write
y13=−i(213×−1−i32){y^{13}} = - i\left( {{2^{13}} \times \dfrac{{ - 1 - i\sqrt 3 }}{2}} \right)
⇒y13=−i[212×(−1−i3)]\Rightarrow {y^{13}} = - i\left[ {{2^{12}} \times \left( { - 1 - i\sqrt 3 } \right)} \right]
⇒y13=212×(−3+i)\Rightarrow {y^{13}} = {2^{12}} \times \left( { - \sqrt 3 + i} \right)
⇒y13=−2123+212i\Rightarrow {y^{13}} = - {2^{12}}\sqrt 3 + {2^{12}}i
⇒Re⁡(y13)=−2123\Rightarrow \operatorname{Re} \left( {{y^{13}}} \right) = - {2^{12}}\sqrt 3
Thus the real [art of the given expression is −2123- {2^{12}}\sqrt 3.
As option ‘b’ corresponds to the answer we obtained. Therefore, it is the correct choice.
So, the correct answer is “Option B”.

Note : While finding out the expansion of a complex expression, always try to reduce the expression to an expression known to us whose power calculation will be easy. Try not to use the binomial expansion as it is lengthy and time consuming.