Question

How do you find the square root of $8i$

Answer 1

Hint : Use de Moivre’s theorem for the square root and such that we get it in principle form which gives the principal root.
First, we are going to find the general polar form for $8i$ and after that we are going to apply root on both sides and then we are going to apply de Moivre’s theorem for the square root and on applying the trigonometric value, we will get the square root of the given and with that we find the other root.

Complete step by step solution:
First, we are going to consider $8i$ and write the polar form for that, which is
$8i = 8(0 + i0) \\\ 8i = 8\left( {\cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)} \right) \;$
Since, we have to find for the square root of the above, we are going to apply square root on both side, which gives us
$\sqrt {8i} = \sqrt {8\left( {\cos \left( {\dfrac{\pi }{2}} \right) + i\sin \left( {\dfrac{\pi }{2}} \right)} \right)}$
Now, we are going to apply de Moivre’s theorem, which is given below
${\left[ {r(\cos \theta + i\sin \theta } \right]^n} = {r^n}\left[ {\cos n\theta + i\sin n\theta } \right]$
On applying the de Moivre’s theorem into the obtained polar form, we will get
$\sqrt {8i} = \sqrt 8 \left( {\cos \left( {\left( {\dfrac{1}{2}} \right)\dfrac{\pi }{2}} \right) + i\sin \left( {\left( {\dfrac{1}{2}} \right)\dfrac{\pi }{2}} \right)} \right) \\\ \sqrt {8i} = \sqrt 8 \left( {\cos \left( {\dfrac{\pi }{4}} \right) + i\sin \left( {\dfrac{\pi }{4}} \right)} \right) \;$
Now, we are going to simplify the RHS, we get

$$= 2\sqrt 2 (\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}i) \\\ = 2 + 2i \;$$

The above obtained is the principal square root. We can also find the other root by multiplying the minus sign to the obtained root. We will get the other root of the given complex number
$- 2 - 2i$

Note : We have to be familiar with the properties of the complex number and also with de Moivre’s theorem as it completely eliminates the complexity of the square root on the complex number and also, we should know the trigonometric value of the necessary functions.