
Find the square root of \[- i\].
Answer
499.2k+ views
Hint: We need to find the square root of \[\ - i\] . Square root of a number is a value in which it turns to the original number when it is multiplied by itself. Suppose \[a\] is a square root of \[b\] then it is represented as \[a\ = \ \sqrt{b}\] . For example, \[3\] is the square root of \[9\] then it is represented as \[3 = \sqrt{9}\]
Complete answer: Given, \[- i\]
We need to find \[\sqrt{- i}\]
Let us assume \[\sqrt{- i} = \ a -{ib}\]
By squaring on both sides ,
\[\left( \sqrt{- i} \right)^{2} = \ \left( a - ib \right)^{2}\]
By expanding the formula,
We get ,
\[i = \ a^{2} + i^{2}b^{2} – 2iab\]
\[i = a^{2} – b^{2} – 2iab\]
By comparing the imaginary part,
\[i = - 2iab\]
By simplifying we get,
\[2ab = - 1\]
\[ab = - \dfrac{1}{2}\]
By comparing the imaginary part,
\[i = - 2iab\]
By simplifying we get,
\[2ab = - 1\]
\[ab = - \dfrac{1}{2}\]
Also by comparing the real part,
\[a^{2} – b^{2} = 0\])
Thus we know that,
\[\left( a^{2} + b^{2} \right)^{2} = \left( a^{2} – b^{2} \right)^{2} + 4a^{2}b^{2}\]
\[\left( a^{2} + b^{2} \right)^{2} = \left( a^{2} – b^{2} \right)^{2} + 4{ab}^2\]
By substituting the known values,
We get,
\[= 0 + 4\left( - \dfrac{1}{2} \right)^{2}\]
\[= 4\left( \dfrac{1}{4} \right)\]
By dividing,
We get,
\[=1\]
Therefore,
The possibilities of \[a\] = \[b =\]
\[\dfrac{1}{\sqrt{2}}\]or\[\ - \dfrac{1}{\sqrt{2}}\]
So by substituting the values in \[\sqrt{- i}\] ,
We get the value of square root of\[\ - i\]
\[\sqrt{- i} = \pm \dfrac{1}{\sqrt{2}} – i\dfrac{1}{\sqrt{2}}\]
By taking the common terms outside,
We get,
\[\sqrt{- i} = \pm \dfrac{1}{\sqrt{2}}\left( 1 – I \right)\]
Thus we got the value.
Final answer :
The value of square root of \[- i\ \] is \[\pm \dfrac{1}{\sqrt{2}}\left( 1 – I \right)\]
Note:
Mathematically, the symbol \[\sqrt{}\] is known as a Radical sign which is used to represent the square root. It is basically one of the methods to solve the quadratic equation . In order to find the square root, we can use two methods
1.Long division method
2.Factorisation method
It’s easy to memorize the square root values of the numbers \[1 to 25\]. After that number we need to use the method to find the values.
Complete answer: Given, \[- i\]
We need to find \[\sqrt{- i}\]
Let us assume \[\sqrt{- i} = \ a -{ib}\]
By squaring on both sides ,
\[\left( \sqrt{- i} \right)^{2} = \ \left( a - ib \right)^{2}\]
By expanding the formula,
We get ,
\[i = \ a^{2} + i^{2}b^{2} – 2iab\]
\[i = a^{2} – b^{2} – 2iab\]
By comparing the imaginary part,
\[i = - 2iab\]
By simplifying we get,
\[2ab = - 1\]
\[ab = - \dfrac{1}{2}\]
By comparing the imaginary part,
\[i = - 2iab\]
By simplifying we get,
\[2ab = - 1\]
\[ab = - \dfrac{1}{2}\]
Also by comparing the real part,
\[a^{2} – b^{2} = 0\])
Thus we know that,
\[\left( a^{2} + b^{2} \right)^{2} = \left( a^{2} – b^{2} \right)^{2} + 4a^{2}b^{2}\]
\[\left( a^{2} + b^{2} \right)^{2} = \left( a^{2} – b^{2} \right)^{2} + 4{ab}^2\]
By substituting the known values,
We get,
\[= 0 + 4\left( - \dfrac{1}{2} \right)^{2}\]
\[= 4\left( \dfrac{1}{4} \right)\]
By dividing,
We get,
\[=1\]
Therefore,
The possibilities of \[a\] = \[b =\]
\[\dfrac{1}{\sqrt{2}}\]or\[\ - \dfrac{1}{\sqrt{2}}\]
So by substituting the values in \[\sqrt{- i}\] ,
We get the value of square root of\[\ - i\]
\[\sqrt{- i} = \pm \dfrac{1}{\sqrt{2}} – i\dfrac{1}{\sqrt{2}}\]
By taking the common terms outside,
We get,
\[\sqrt{- i} = \pm \dfrac{1}{\sqrt{2}}\left( 1 – I \right)\]
Thus we got the value.
Final answer :
The value of square root of \[- i\ \] is \[\pm \dfrac{1}{\sqrt{2}}\left( 1 – I \right)\]
Note:
Mathematically, the symbol \[\sqrt{}\] is known as a Radical sign which is used to represent the square root. It is basically one of the methods to solve the quadratic equation . In order to find the square root, we can use two methods
1.Long division method
2.Factorisation method
It’s easy to memorize the square root values of the numbers \[1 to 25\]. After that number we need to use the method to find the values.
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