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Question: Solve \[4 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac...

Solve 4+5(12+i32)334+3(12+i32)3654 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} where, i=1i = \sqrt { - 1} .
(a) 11
(b) i2i\sqrt 2
(c) ii
(d) 3 i\sqrt 3 {\text{ }}i

Explanation

Solution

We will be going to use the most curious concept of complex relations to be recognised by those three different formulae of “cube root of unity” containing omega and its respective value, for the analysis of desired expression.

Complete step by step answer:
\because The condition is related to the complex number as there exists the parameter ‘i’ where the value of instance ‘i’ is i=1i = \sqrt { - 1} respectively.
Here, we have given the expression as 4+5(12+i32)334+3(12+i32)3654 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}}
So, pre-assuming the term(12+i32)=ω\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right) = \omega , this is the complex term.
The given equation becomes,
4+5(12+i32)334+3(12+i32)365=4+5ω334+3ω3354 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 4 + 5{\omega ^{334}} + 3{\omega ^{335}}
Therefore, using the complex algebraic relation for solution we can reach to the desired value.
Since, using the rules of indices, adjusting the powers so that we get the respective output as per the terminology,
4+5(12+i32)334+3(12+i32)365=4+5(ω3)111ω+3(ω3)121ω24 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 4 + 5{\left( {{\omega ^3}} \right)^{111}}\omega + 3{\left( {{\omega ^3}} \right)^{121}}{\omega ^2} 4+5(12+i32)334+3(12+i32)365=4+5ω+3ω24 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 4 + 5\omega + 3{\omega ^2}
\because We know that, ω3=1{\omega ^3} = 1
Now, since expanding the terms to the desired formula, we get
4+5(12+i32)334+3(12+i32)365=1+3+2ω+3ω+3ω24 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 1 + 3 + 2\omega + 3\omega + 3{\omega ^2}
Rearranging the terms to the efficient grouping, we get
4+5(12+i32)334+3(12+i32)365=1+2ω+3+3ω+3ω24 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 1 + 2\omega + 3 + 3\omega + 3{\omega ^2}
Now, taking the common terms, we get
4+5(12+i32)334+3(12+i32)365=1+2ω+3(1+ω+ω2)4 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 1 + 2\omega + 3(1 + \omega + {\omega ^2})
Substituting the desire value in place of respective formula formed that is, 1+ω+ω2=01 + \omega + {\omega ^2} = 0, we get
4+5(12+i32)334+3(12+i32)365=1+2ω+04 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 1 + 2\omega + 0
Now, replacing the value of omega in place of the solution that we have assumed previously in the solution, we get
4+5(12+i32)334+3(12+i32)365=1+2(12+i32)4 + 5{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)^{365}} = 1 + 2\left( { - \dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)
Simplifying the above equation after putting the assumed value, we get
4+5(12+i32)334+3(12+i32)365=1+1+i34 + 5{\left( { - \dfrac{1}{2} + i\sqrt {\dfrac{3}{2}} } \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\sqrt {\dfrac{3}{2}} } \right)^{365}} = 1 + - 1 + i\sqrt 3
4 + 5{\left( { - \dfrac{1}{2} + i\sqrt {\dfrac{3}{2}} } \right)^{334}} + 3{\left( { - \dfrac{1}{2} + i\sqrt {\dfrac{3}{2}} } \right)^{365}} = i\sqrt 3 $$$$ = \sqrt 3 i
\therefore The required answer is 1 - 1. The correct option is (b).

Note:
Considering the three cube roots of unity i.e. 11,ω\omega and ω2{\omega ^2} where, ω=1+i32\omega = \dfrac{{ - 1 + i\sqrt 3 }}{2} and ω2=1i32{\omega ^2} = \dfrac{{ - 1 - i\sqrt 3 }}{2}. Since, we also know that, 1+ω+ω2=01 + \omega + {\omega ^2} = 0 and ω3=1{\omega ^3} = 1. As a result, use the respective formulae and conditions in the calculations to find the desired output (before starting the solutions assume 1+i32\dfrac{{ - 1 + i\sqrt 3 }}{2} as ω\omega . Use the proper conditions for the indices and algebraic rules for simplification of expression.