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Question: Evaluate the value of \({{\left( -\sqrt{-1} \right)}^{4n+3}}\) if \(n\in N\)....

Evaluate the value of (−−1)4n+3{{\left( -\sqrt{-1} \right)}^{4n+3}} if n∈Nn\in N.

Explanation

Solution

Hint: Just solve the question using normal algebra of exponents and small imaginary concepts like i2=−1,i4=1{{i}^{2}}=-1,{{i}^{4}}=1. Substitute the imaginary number into the equation and then solve the complex equation.

Complete step-by-step solution -
Definition of i, can be written as:
The solution of the equation: x2+1=0{{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 numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: (1+i)x+(1+i)=0,x=−1\left( 1+i \right)x+\left( 1+i \right)=0,x= -1 is the root of the equation.
Given expression of n in the question is in the form:
(−−1)4n+3{{\left( -\sqrt{-1} \right)}^{4n+3}}
Substitute the imaginary number i into the expression inside the bracket as the imaginary number i is a solution of second degree equation of xx :
x2+1=0⇒i=−1{{x}^{2}}+1=0\Rightarrow i=\sqrt{-1}
By substituting this our given expression takes the form of:
(−i)4n+3{{\left( -i \right)}^{4n+3}}
Now use a basic concept of exponents in algebra. The identity:
ab+c=ab.ac{{a}^{b+c}}={{a}^{b}}.{{a}^{c}}
By using the above identity our expression takes form of:
(−i)4n+3=(−i)4n.(−i)3{{\left( -i \right)}^{4n+3}}={{\left( -i \right)}^{4n}}.{{\left( -i \right)}^{3}}
Now using the basic identity of imaginary numbers. The identity:
i4=1{{i}^{4}}=1
By using the above identity, we get
=(−1)4n.(i)4n(−1)3(i)3={{\left( -1 \right)}^{4n}}.{{\left( i \right)}^{4n}}{{\left( -1 \right)}^{3}}{{\left( i \right)}^{3}}
By using the basic identity of exponents here. The identity:
abc=(ab)c{{a}^{bc}}={{\left( {{a}^{b}} \right)}^{c}}
=(−1)4n(1)n(−1)(i)3={{\left( -1 \right)}^{4n}}{{\left( 1 \right)}^{n}}\left( -1 \right){{\left( i \right)}^{3}}
Now using basic identity if imaginary numbers. The identity:
i3=−i =1.1.(−1).(−i) \begin{aligned} & {{i}^{3}}=-i \\\ & =1.1.\left( -1 \right).\left( -i \right) \\\ \end{aligned}
By simplifying the above, we get:
(−−1)4n+3=i{{\left( -\sqrt{-1} \right)}^{4n+3}}=i
Then ‘i’ is the value of the required expression.

Note: Be careful while separating terms as if you forgot a minus sign you get -i as answer which is wrong. Here we need to remember the general form of iota and its peoperty.