Question

Find the square root of $ -i $
A. $ \pm \dfrac{1}{\sqrt{2}}\left( 1+i \right) $
B. $ \pm \dfrac{1}{\sqrt{2}}\left( 1-i \right) $
C. $ \pm \dfrac{1}{2}\left( 1+i \right) $
D. $ \pm \dfrac{1}{2}\left( 1-i \right) $

Answer 1

Hint: We will take the given complex number as $ x+iy $ and its square root as $ a+ib $ , so that it can be represented as, $ a+ib=\sqrt{x+iy} $ . On squaring and comparing, we can see that the value of x is $ {{a}^{2}}-{{b}^{2}} $ and y is $ 2ab $ . The complex number given here is $ -i $ , so x, y is 0 and - 1. We will then find the value of a and b from that.

Complete step-by-step answer:
In the question, we have been given a complex number $ -i $ and we have been asked to find its square root. The square root of any complex number is also a complex number. If we take a complex number as $ x+iy $ , and its square root as $ a+ib $ , then it can be represented as, $ a+ib=\sqrt{x+iy} $ . Here a, b, x, y represent real numbers. Now, let us square both sides of the equation. So, we get,
 $ {{\left( a+ib \right)}^{2}}=x+iy $
We know that $ {{\left( a+ib \right)}^{2}} $ can be expanded by using the formula, $ {{\left( c+d \right)}^{2}}={{c}^{2}}+{{d}^{2}}+2cd $ . So, applying this formula, we will expand the above equation. So, we get,
 $ {{a}^{2}}+{{i}^{2}}{{b}^{2}}+2iab=x+iy $
We know that the value of $ {{i}^{2}}=-1 $ . So, we can rewrite the above equation as,
 $ {{a}^{2}}-{{b}^{2}}+2iab=x+iy $
Now, on comparing, we can say that the value of x can be represented as $ {{a}^{2}}-{{b}^{2}} $ and y as $ 2ab $ . So, we have $ x+iy $ and we have to find the value of $ a+ib $ . Also, the complex number has been given here as $ -i $ . So, we have,
 $ x+iy=-i $
Hence, we can say that x is 0 and y is - 1. Now, we also know that x is equal to $ {{a}^{2}}-{{b}^{2}} $ and y is equal to $ 2ab $ . So, we can say that,
 $ {{a}^{2}}-{{b}^{2}}=0,2ab=-1 $
So, we can say from the equation, $ {{a}^{2}}-{{b}^{2}}=0 $ , that, $ {{a}^{2}}={{b}^{2}} $ or a is equal to b or - b.
So, if we take the first case, a = b, then the equation $ 2ab=-1 $ can be written as $ 2{{b}^{2}}=-1 $ , which is not possible as b is a real number.
Now, for the second case, a = - b, the equation $ 2ab=-1 $ can be written as $ -2{{b}^{2}}=-1\text{ }or\text{ }{{b}^{2}}=\dfrac{1}{2}\text{ }or\text{ }b=\pm \dfrac{1}{\sqrt{2}} $ . So, since a = - b, we can say that $ a=\mp \dfrac{1}{\sqrt{2}} $ . Therefore, the square root of - i is $ \mp \dfrac{1}{\sqrt{2}}\pm \dfrac{1}{\sqrt{2}} $ or $ \pm \dfrac{1}{\sqrt{2}}\left( -1+i \right)\text{ }or\text{ }\pm \dfrac{1}{\sqrt{2}}\left( 1-i \right) $.

Hence, the correct answer is option B.
Note: We can also the formula, that if $ Z=-{{r}^{2}}i $ , where r represents any real number, then the value of $ \sqrt{Z}=\pm \dfrac{r}{\sqrt{2}}\left( 1-i \right) $ .