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 $\sqrt{Z}=\pm \dfrac{r}{\sqrt{2}}\left( 1-i \right)$ .