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# What is the square root of i, where $i=\sqrt{-1}$ ?a.$\dfrac{1+i}{2}$b.$\dfrac{1-i}{2}$c.$\dfrac{1+i}{\sqrt{2}}$d.None of the above. Verified
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Hint: First convert ‘I’ into polar form by using the formulae $\tan \alpha =\dfrac{b}{a}$, ${{r}^{2}}={{a}^{2}}+{{b}^{2}}$, and $z=r\times \left( \cos \alpha +i\sin \alpha \right)$ . Then take square roots on both sides and use the Demoivre's theorem i.e. ${{z}^{n}}={{r}^{n}}\times \left[ \cos \left( n\times \alpha \right)+i\sin \left( n\times \alpha \right) \right]$ to get the square root of ‘i’.

To find the value of ‘I’ we will assume it as ‘z’,
$\therefore z=i$
Above equation can also be written as,
$\therefore z=0+i$
If we compare above equation with z = a + bi we can write,
a = 0 , b = 1 ………………………………………………… (i)
Now to find the square root of ‘I’ we have to convert it into polar form so that we can do some operations on that, and for that we have to find the angle $\alpha$ and radius ‘r’ by using the formulae given below,
Formulae:
$\tan \alpha =\dfrac{b}{a}$
${{r}^{2}}={{a}^{2}}+{{b}^{2}}$
By using formula (1) and values of equation (i) we will get,
$\therefore \tan \alpha =\dfrac{1}{0}$
As we know the value of $\dfrac{1}{0}$ is $\infty$ therefore we will get,
$\therefore \tan \alpha =\infty$
$\therefore \alpha ={{\tan }^{-1}}\infty$
As, $\tan \dfrac{\pi }{2}=\infty$ therefore ${{\tan }^{-1}}\infty =\dfrac{\pi }{2}$, therefore above equation will become,
$\therefore \alpha =\dfrac{\pi }{2}$ ……………………………………………… (ii)
Also, By using formula and the values of equation (i) we will get,
${{r}^{2}}={{a}^{2}}+{{b}^{2}}$
$\therefore {{r}^{2}}={{0}^{2}}+{{1}^{2}}$
$\therefore {{r}^{2}}={{1}^{2}}$
Taking square root on both sides of the equation we will get,
$\therefore r=1$ ……………………………………………… (iii)
As we have to convert ‘i’ in to polar form therefore we should know the formula of polar form given below,
Formula:
$z=r\times \left( \cos \alpha +i\sin \alpha \right)$
If we put the values of equation (ii) and equation (iii) in the above equation we will get,
$\therefore z=1\times \left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)$
$\therefore z=\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)$
As we assumed z = i therefore above equation will become,
$\therefore i=\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)$
As we have to find the square root to ‘i’ therefore we will take square roots on both sides of the above equation,
$\therefore \sqrt{i}=\sqrt{\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)}$
As we all know that the square root is nothing but the $\dfrac{1}{2}$th power of the term, therefore above equation will become,
$\therefore \sqrt{i}={{\left( \cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2} \right)}^{\dfrac{1}{2}}}$ ……………………………………………….. (iv)
Now to proceed further in the solution we should know the Demoivre's theorem given below,
Demoivre's theorem:
If ‘z’ be any complex number represented by $z=r\times \left( \cos \alpha +i\sin \alpha \right)$ then,
${{z}^{n}}={{r}^{n}}\times \left[ \cos \left( n\times \alpha \right)+i\sin \left( n\times \alpha \right) \right]$
By using Demoivre's theorem, equation (iv) will become,
$\therefore \sqrt{i}=\cos \left( \dfrac{1}{2}\times \dfrac{\pi }{2} \right)+i\sin \left( \dfrac{1}{2}\times \dfrac{\pi }{2} \right)$
$\therefore \sqrt{i}=\cos \left( \dfrac{\pi }{4} \right)+i\sin \left( \dfrac{\pi }{4} \right)$
As we know that, $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ and $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ therefore above equation will become,
$\therefore \sqrt{i}=\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}$
$\therefore \sqrt{i}=\dfrac{1+i}{\sqrt{2}}$
Therefore the value of $\sqrt{i}$ is equal to $\dfrac{1+i}{\sqrt{2}}$.
Therefore the correct answer is option (c)

Note: Demoivre's theorem i.e. ${{z}^{n}}={{r}^{n}}\times \left[ \cos \left( n\times \alpha \right)+i\sin \left( n\times \alpha \right) \right]$ plays very important role in this problem as without using it you can face a very lengthy solution therefore to avoid the complexity try to remember it.
Last updated date: 25th Sep 2023
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