Question

If n is a positive number but not a multiple of 3 and $z=-1+i\sqrt{3},\left( where\text{ }i=\sqrt{-1} \right)$ then $\left( {{z}^{2n}}+{{2}^{n}}{{z}^{n}}+{{2}^{2n}} \right)$ is equal to
(a) 0
(b) 1
(c) 1
(d) $3\times {{2}^{n}}$

Answer 1

Find the conditions possible on n. As given n is an integer the cases possible are n is odd and n is even. So, check the values in both the cases. Given n is not multiple of 3. So the possible remainder to be 1 and 2. So we can write n as 3k+1 and 3k+2. Solve for these both cases.

Complete step-by-step solution:
Definition of i, can be written as:
The solution of the equation: ${{x}^{2}}+1=0$ is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as: An equation containing complex number in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0,x= -1$ is the root of the equation.
By looking at the given expression in the question is:
${{z}^{2n}}+{{2}^{n}}{{z}^{n}}+{{2}^{2n}}$
By directly simplifying the value of z we get as:
$z=-1+i\sqrt{3}$
By multiplying and dividing with 2 on z, we get:
$z=-\dfrac{1+i\sqrt{3}}{2}\times 2=\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)\times 2$
By using Euler’s formula of cos and sin of angles, we get:
$cisx=\cos x+i\sin x={{e}^{ix}}$
By substituting Euler’s formula of cos and sin into expression, we get:
$\begin{aligned} & x=\dfrac{2\pi }{3} \\\ & z=\left( {{e}^{i\dfrac{2\pi }{3}}} \right)2 \\\ \end{aligned}$
By substituting this solution of z into the expression we get:
${{\left( 2{{e}^{i\dfrac{2\pi }{3}}} \right)}^{2n}}+{{2}^{n}}{{\left( 2{{e}^{i\dfrac{2\pi }{3}}} \right)}^{n}}+{{2}^{2n}}$
By simplifying the above equation of z into the expression we get
$={{2}^{2n}}{{e}^{i\dfrac{4\pi n}{3}}}+{{2}^{2n}}{{e}^{i\dfrac{2\pi n}{3}}}+{{2}^{2n}}$
By taking the term which is common in the expression we get:
$={{2}^{2n}}\left( {{e}^{i\dfrac{4\pi n}{3}}}+{{e}^{i\dfrac{2\pi n}{3}}}+1 \right)$
Given the condition in the question on n is n is not multiple of 3.
n = 3k + 1 (or) n = 3k + 2
In the expression we can write $4\pi n=3\pi n+\pi n$
By substituting in the above we get
$={{2}^{2n}}\left( {{e}^{i\pi n}} . {{e}^{i\dfrac{\pi n}{3}}}+{{e}^{{{e}^{i\dfrac{2\pi n}{3}}}}}+1 \right)$
Case 1: n = 3k + 1
$={{2}^{2n}}\left( {{e}^{i\dfrac{\left( 12k+4 \right)\pi }{3}}}+{{e}^{i\dfrac{\left( 6k+2 \right)\pi }{3}}}+1 \right)$
By simplifying the above equation, we get:
$={{2}^{2n}}\left( {{e}^{4k\pi i}}.{{e}^{\dfrac{4\pi }{3}i}}+{{e}^{i\left( 2k\pi \right)}}.{{e}^{i\dfrac{2\pi }{3}}}+1 \right)$
We know the value of sin ($2k\pi$) and sin($4k\pi$) is 0.
By substituting this and then converting it into the terms of sin and cos, we get:
$={{2}^{2n}}\left( \left( \cos \left( 4k\pi \right) \right).\left( \cos \left( \dfrac{4\pi }{3} \right)+i\left( \sin \left( \dfrac{4\pi }{3} \right) \right) \right)+\left( \cos \left( 2k\pi \right) \right).\left( \cos \left( \dfrac{2\pi }{3} \right)+i\left( \sin \left( \dfrac{2\pi }{3} \right) \right) \right)+1 \right)$
We know values as:

& \cos \left( 4k\pi \right)=1 \\\ & \cos \left( \dfrac{4\pi }{3} \right)=-\dfrac{1}{2} \\\ & \sin \left( \dfrac{4\pi }{3} \right)=-\dfrac{\sqrt{3}}{2} \\\ & \cos \left( 2k\pi \right)=1 \\\ & \cos \left( \dfrac{2\pi }{3} \right)=-\dfrac{1}{2} \\\ & \sin \left( \dfrac{2\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$$ By substituting these into equation we get: $={{2}^{2n}}\left( 1+\dfrac{-1}{2}+\dfrac{-\sqrt{3}}{2}i+\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}i \right)$ By cancelling common terms we get expression value to be 0. Case 2: n = 3k + 2 $={{2}^{2n}}\left( 1+{{e}^{i\dfrac{8\pi }{3}}}+{{e}^{i\dfrac{4\pi }{3}}} \right)$ By simplifying above equation, we get: $={{2}^{2n}}\left( 1+\left( \cos \left( \dfrac{8\pi }{3} \right)+i\left( \sin \left( \dfrac{8\pi }{3} \right) \right) \right)+\left( \cos \left( \dfrac{4\pi }{3} \right)+i\left( \sin \left( \dfrac{4\pi }{3} \right) \right) \right) \right)$ By basic knowledge of trigonometry, we know the values : $$\begin{aligned} & \cos \left( \dfrac{4\pi }{3} \right)=-\dfrac{1}{2} \\\ & \sin \left( \dfrac{4\pi }{3} \right)=-\dfrac{\sqrt{3}}{2} \\\ & \cos \left( \dfrac{8\pi }{3} \right)=-\dfrac{1}{2} \\\ & \sin \left( \dfrac{8\pi }{3} \right)=\dfrac{\sqrt{3}}{2} \\\ \end{aligned}$$ By substituting the above values into expressions all terms get canceled we get value to be 0. So, 0 is the value of expression in any case. **Option (a) is correct.** **Note:** Be careful while taking cases for n and while converting it into terms of sin and cos as it defines the whole answer. After converting into sin and cos observe the values which can be canceled. In terms of cis x the value directly turns into 1 if x is even integral multiple of $$\pi $$. 2k and 4k are both even.