Question

The modulus of $\sqrt {2i} - \sqrt { - 2i} $ is
$
  (a){\text{ 2}} \\
  (b){\text{ }}\sqrt 2 \\
  (c){\text{ 0}} \\
  (d){\text{ 2}}\sqrt 2 \\
$

Answer 1

Hint: In this question write the negative sign present inside the square root as $i$ because $i = \sqrt { - 1} $. Then take $\sqrt i $ common, and substitute it as $\sqrt i = a + ib$, simple algebraic manipulations will help evaluation of the given complex number.

Complete step-by-step answer:
Given expression
$\sqrt {2i} - \sqrt { - 2i} $
Now as we know in complex $\left[ { - 1 = {i^2}} \right]$
$ \Rightarrow \sqrt {2i} - \sqrt { - 2i} = \sqrt {2i} - \sqrt {{i^2}2i} $
$ \Rightarrow \sqrt {2i} - \sqrt { - 2i} = \sqrt {2i} - \sqrt {{i^2}2i} = \sqrt {2i} - i\sqrt {2i} = \sqrt 2 \sqrt i \left( {1 - i} \right)$..................... (a)
Now let$\sqrt i = a + ib$............................ (1)
Squaring on both sides we have,
$i = {\left( {a + ib} \right)^2} = {a^2} + {i^2}{b^2} + 2iab = {a^2} - {b^2} + 2iab$
Now compare real and imaginary terms we have,
$ \Rightarrow {a^2} - {b^2} = 0$............................. (2)
And $2ab = 1$.............................. (3)
Now from equation (2) we have
$ \Rightarrow {a^2} = {b^2}$
$ \Rightarrow a = b$....................... (4)
Now from equation (3) we have,
$ \Rightarrow 2{a^2} = 1$
$ \Rightarrow a = \pm \dfrac{1}{{\sqrt 2 }}$
Now from equation (4) we have,
$ \Rightarrow a = b = \pm \dfrac{1}{{\sqrt 2 }}$
Now from equation (1) we have,
$ \Rightarrow \sqrt i = \pm \dfrac{1}{{\sqrt 2 }}\left( {1 + i} \right)$
Now from equation (a) we have,
$ \Rightarrow \sqrt {2i} - \sqrt { - 2i} = \sqrt 2 \sqrt i \left( {1 - i} \right) = \sqrt 2 \times \left( { \pm \dfrac{1}{{\sqrt 2 }}\left( {1 + i} \right)} \right)\left( {1 - i} \right)$
$ \Rightarrow \sqrt {2i} - \sqrt { - 2i} = \pm \left( {1 + i} \right)\left( {1 - i} \right) = \pm \left( {1 - {i^2}} \right)$
$\left[ {\because - 1 = {i^2}} \right]$
$ \Rightarrow \sqrt {2i} - \sqrt { - 2i} = \pm \left( {1 - \left( { - 1} \right)} \right) = \pm 2$
Now we have to find out the modulus of $\sqrt {2i} - \sqrt { - 2i} $
$ \Rightarrow \left| {\sqrt {2i} - \sqrt { - 2i} } \right| = \left| { \pm 2} \right| = 2$
So this is the required answer.
Hence option (A) is correct.

Note: It’s important to understand the physical significance of the modulus of a complex number. It is the length of the directed line segment drawn from the origin of the complex plane to the point (a, b) where a is the real part and b is the imaginary part of any complex number a+ ib.