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Question: How do you find the two square roots of \(2i\)....

How do you find the two square roots of 2i2i.

Explanation

Solution

In this question we will consider the square root of the term as some variable zz and then use the exponential form of complex numbers to simplify the expression and then use Euler’s formula and simplify to get the required solution.

Formula used: Exponential form of complex number: z=reiθz = r{e^{i\theta }}
Euler’s formula: eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta

Complete step-by-step solution:
We have the given terms as:2i2i
And we have to find the square root of this term therefore, we can write it as:
⇒2i\Rightarrow \sqrt {2i}
For simplification purposes let’s consider:
⇒z=2i\Rightarrow z = \sqrt {2i}
Now on squaring both the sides, we get:
⇒z2=(2i)2\Rightarrow {z^2} = {\left( {\sqrt {2i} } \right)^2}
On taking the square on both the sides, we get:
⇒z2=2i\Rightarrow {z^2} = 2i
Now on using the exponential form of complex numbers we can write:
⇒2i=r2ei(2θ)\Rightarrow 2i = {r^2}{e^{i(2\theta )}}
⇒r2=2\Rightarrow {r^2} = 2 which indicates that the value of r=2r = \sqrt 2
Now 2θ=π2+2nπ2\theta = \dfrac{\pi }{2} + 2n\pi therefore, θ=π4+nπ\theta = \dfrac{\pi }{4} + n\pi
Therefore, we can write the value of the term as:
⇒z=2ei(π2+nπ)\Rightarrow z = \sqrt 2 {e^{i\left( {\dfrac{\pi }{2} + n\pi } \right)}}
Now on using Euler’s formula on the above expression, we get:
⇒z=2[cos⁡(π4+nπ)+isin⁡(π4+nπ)]\Rightarrow z = \sqrt 2 \left[ {\cos \left( {\dfrac{\pi }{4} + n\pi } \right) + i\sin \left( {\dfrac{\pi }{4} + n\pi } \right)} \right]
Now we know that cos⁡π4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} and it toggles to the negative value at every π\pi radians therefore the value of cos⁡(π4+nπ)\cos \left( {\dfrac{\pi }{4} + n\pi } \right) is ±12 \pm \dfrac{1}{{\sqrt 2 }}
And we also know that that sin⁡π4=12\sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} and it toggles to the negative value at every π\pi radians therefore the value of sin⁡(π4+nπ)\sin \left( {\dfrac{\pi }{4} + n\pi } \right) is ±12 \pm \dfrac{1}{{\sqrt 2 }}
Therefore, on substituting it in the above expression, we get:
z=2(±12±i12)z = \sqrt 2 \left( { \pm \dfrac{1}{{\sqrt 2 }} \pm i\dfrac{1}{{\sqrt 2 }}} \right)
On taking the common denominator on both the terms, we get:
z=2(±1±i2)z = \sqrt 2 \left( {\dfrac{{ \pm 1 \pm i}}{{\sqrt 2 }}} \right)
On simplifying, we get:
z=±1±iz = \pm 1 \pm i

Therefore 2i2i has 22 square roots, which are 1+i1 + i and −1−i - 1 - i, which are the required solution.

Note: For doing questions on complex numbers, the polar format and the exponential format of the complex number should be remembered.
It is to be remembered that when both the sides of the expression are squared, the value of the expression does not change.