Question

If $i=\sqrt{-1}$, then calculate the value of $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}$.
(a) $1-i\sqrt{3}$
(b) $-1+i\sqrt{3}$
(c) $i\sqrt{3}$
(d) $-i\sqrt{3}$

Answer 1

Hint: Use the fact that $w=\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}$ is a cube root of unity and thus ${{w}^{3}}=1$ . Simplify the given expression, use the laws of exponents which state that ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$ and ${{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}$ and calculate the value of the expression.

Complete step-by-step solution -
We have to calculate the value of the expression $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}$.
We know that $w=\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}$ is a cube root of unity. Thus, we have ${{w}^{3}}=1$.
So, we can rewrite the expression $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}$ as $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5{{w}^{334}}+3{{w}^{365}}$.
We know that the laws of exponents state that ${{a}^{b}}\times {{a}^{c}}={{a}^{b+c}}$ and ${{\left( {{a}^{b}} \right)}^{c}}={{a}^{bc}}$.
So, we can rewrite ${{w}^{334}}$ as ${{w}^{334}}={{w}^{333+1}}={{\left( {{w}^{3}} \right)}^{111}}\times w$.
Similarly, we can rewrite ${{w}^{365}}$ as ${{w}^{365}}={{w}^{363+2}}={{\left( {{w}^{3}} \right)}^{121}}\times {{w}^{2}}$.
Substituting these values in the expression $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5{{w}^{334}}+3{{w}^{365}}$, we have $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5{{w}^{334}}+3{{w}^{365}}=4+5\left[ {{\left( {{w}^{3}} \right)}^{111}}\times w \right]+3\left[ {{\left( {{w}^{3}} \right)}^{121}}\times {{w}^{2}} \right]$.
We know that ${{w}^{3}}=1$.
Thus, we have $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5{{w}^{334}}+3{{w}^{365}}=4+5\left[ {{1}^{111}}\times w \right]+3\left[ {{1}^{121}}\times {{w}^{2}} \right]=4+5w+3{{w}^{2}}$.
By rearranging the terms of the above equation, we have $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5w+3{{w}^{2}}=4+4w+4{{w}^{2}}+w-{{w}^{2}}$.
We know that the roots of the equation ${{x}^{3}}=1$ are $1,w,{{w}^{2}}$.
We also know that if $\alpha ,\beta ,\gamma $ are roots of the cubic equation of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d$, then we have $\alpha +\beta +\gamma =\dfrac{-b}{a}$, $\alpha \beta +\beta \gamma +\gamma \alpha =\dfrac{c}{a}$ and $\alpha \beta \gamma =\dfrac{-d}{a}$.
Thus, we have $1+w+{{w}^{2}}=\dfrac{0}{1}=0$.
So, we have $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5w+3{{w}^{2}}=4+4w+4{{w}^{2}}+w-{{w}^{2}}=4\left( 1+w+{{w}^{2}} \right)+w-{{w}^{2}}$ .
Thus, we have $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4\left( 1+w+{{w}^{2}} \right)+w-{{w}^{2}}=4\left( 0 \right)+w-{{w}^{2}}=w-{{w}^{2}}$.
We will now calculate the value of ${{w}^{2}}$ using the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
As $w=\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}$, we have ${{w}^{2}}={{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}+{{\left( \dfrac{i\sqrt{3}}{2} \right)}^{2}}+2\left( \dfrac{-1}{2} \right)\left( \dfrac{i\sqrt{3}}{2} \right)$.
As $i=\sqrt{-1}$, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$.
Thus, we have ${{w}^{2}}={{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{2}}={{\left( \dfrac{-1}{2} \right)}^{2}}+{{\left( \dfrac{i\sqrt{3}}{2} \right)}^{2}}+2\left( \dfrac{-1}{2} \right)\left( \dfrac{i\sqrt{3}}{2} \right)=\dfrac{1}{4}+\dfrac{\left( -1 \right)3}{4}-\dfrac{i\sqrt{3}}{2}=\dfrac{1-3}{4}-\dfrac{i\sqrt{3}}{2}=\dfrac{-1}{2}-\dfrac{i\sqrt{3}}{2}$.
Substituting $w=\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}$ and ${{w}^{2}}=\dfrac{-1}{2}-\dfrac{i\sqrt{3}}{2}$ in the equation $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4\left( 1+w+{{w}^{2}} \right)+w-{{w}^{2}}=4\left( 0 \right)+w-{{w}^{2}}=w-{{w}^{2}}$, we have $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=w-{{w}^{2}}=\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}-\left( \dfrac{-1}{2}-\dfrac{i\sqrt{3}}{2} \right)=\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}+\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}=i\sqrt{3}$ .
Hence, the value of the expression $4+5{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( \dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}$ is $i\sqrt{3}$, which is option (c).

Note: We can’t solve this question without using the fact that $1,w,{{w}^{2}}$ are roots of the equation ${{x}^{3}}=1$. We also need to use the law of exponents to simplify the powers of the exponents; otherwise, it will be very time consuming to solve this question.