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Question

Question: What is the value of \({{i}^{-1}}\) ?...

What is the value of i1{{i}^{-1}} ?

Explanation

Solution

The given question is solved using the concept of complex numbers. Firstly, we multiply and divide the given expression with the imaginary number ii . Then, we substitute the value of i2{{i}^{2}}to get the required result.

Complete step-by-step solution:
We are given an expression in the question and need to simplify it. We will be using the concept of complex numbers to simplify the expression.
Complex numbers, in mathematics, are represented in the form p+iqp+iq
Here,
p, q are the real numbers
ii is the unit imaginary number
The unit imaginary number ii is equal to the square root of minus one.
The value of ii is given by,
i=1\Rightarrow i=\sqrt{-1}
Squaring the above equation on both sides, we get,
i2=1\Rightarrow {{i}^{2}}=-1
The inverse of the given number is the reverse of the number itself.
For example,
The inverse of the number aa is 1a\dfrac{1}{a}
The value 1a\dfrac{1}{a} is also represented as a1{{a}^{-1}}
From the above,
a1=1a\Rightarrow {{a}^{-1}}=\dfrac{1}{a}
Applying the same for i1{{i}^{-1}} , we get,
i1=1i\Rightarrow {{i}^{-1}}=\dfrac{1}{i}
Multiplying and dividing with an imaginary number ii on the right-hand side, we get,
i1=1i×ii\Rightarrow {{i}^{-1}}=\dfrac{1}{i}\times \dfrac{i}{i}
Multiplying the denominator on the right-hand side, we get,
i1=ii2\Rightarrow {{i}^{-1}}=\dfrac{i}{{{i}^{2}}}
From the above, we know that the value of i2=1{{i}^{2}}=-1
Substituting the value of i2{{i}^{2}}, we get,
i1=i(1)\Rightarrow {{i}^{-1}}=\dfrac{i}{\left( -1 \right)}
Simplifying the above expression, we get,
i1=i\therefore {{i}^{-1}}=-i
The value of the inverse of the imaginary number ii is equal to the negative of the imaginary number ii or i-i

Note: The question can also be solved using the concept of exponents as follows,
The expression i1{{i}^{-1}} is multiplied and divided by i2{{i}^{2}}
i1×i2i2\Rightarrow {{i}^{-1}}\times \dfrac{{{i}^{2}}}{{{i}^{2}}}
According to the rules of exponents,
am×an=a(m+n)\Rightarrow {{a}^{m}}\times {{a}^{n}}={{a}^{\left( m+n \right)}}
Applying the same, we get,
i1+2i2\Rightarrow \dfrac{{{i}^{-1+2}}}{{{i}^{2}}}
Simplifying the above expression,
ii2\Rightarrow \dfrac{i}{{{i}^{2}}}
Substituting the value of i2{{i}^{2}} ,
i(1)\Rightarrow \dfrac{i}{\left( -1 \right)}
i1=i\therefore {{i}^{-1}}=-i