Solveeit Logo
Login

Question

Question: The modulus of \(\sqrt {2i} - \sqrt { - 2i} \) is \( (a){\text{ 2}} \\\ (b){\text{ }}\sqrt...

The modulus of 2i−−2i\sqrt {2i} - \sqrt { - 2i} is
(a) 2 (b) 2 (c) 0 (d) 22  (a){\text{ 2}} \\\ (b){\text{ }}\sqrt 2 \\\ (c){\text{ 0}} \\\ (d){\text{ 2}}\sqrt 2 \\\

Explanation

Solution

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

Complete step-by-step answer:
Given expression
2i−−2i\sqrt {2i} - \sqrt { - 2i}
Now as we know in complex [−1=i2]\left[ { - 1 = {i^2}} \right]
⇒2i−−2i=2i−i22i\Rightarrow \sqrt {2i} - \sqrt { - 2i} = \sqrt {2i} - \sqrt {{i^2}2i}
⇒2i−−2i=2i−i22i=2i−i2i=2i(1−i)\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 leti=a+ib\sqrt i = a + ib............................ (1)
Squaring on both sides we have,
i=(a+ib)2=a2+i2b2+2iab=a2−b2+2iabi = {\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,
⇒a2−b2=0\Rightarrow {a^2} - {b^2} = 0............................. (2)
And 2ab=12ab = 1.............................. (3)
Now from equation (2) we have
⇒a2=b2\Rightarrow {a^2} = {b^2}
⇒a=b\Rightarrow a = b....................... (4)
Now from equation (3) we have,
⇒2a2=1\Rightarrow 2{a^2} = 1
⇒a=±12\Rightarrow a = \pm \dfrac{1}{{\sqrt 2 }}
Now from equation (4) we have,
⇒a=b=±12\Rightarrow a = b = \pm \dfrac{1}{{\sqrt 2 }}
Now from equation (1) we have,
⇒i=±12(1+i)\Rightarrow \sqrt i = \pm \dfrac{1}{{\sqrt 2 }}\left( {1 + i} \right)
Now from equation (a) we have,
⇒2i−−2i=2i(1−i)=2×(±12(1+i))(1−i)\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)
⇒2i−−2i=±(1+i)(1−i)=±(1−i2)\Rightarrow \sqrt {2i} - \sqrt { - 2i} = \pm \left( {1 + i} \right)\left( {1 - i} \right) = \pm \left( {1 - {i^2}} \right)
[∵−1=i2]\left[ {\because - 1 = {i^2}} \right]
⇒2i−−2i=±(1−(−1))=±2\Rightarrow \sqrt {2i} - \sqrt { - 2i} = \pm \left( {1 - \left( { - 1} \right)} \right) = \pm 2
Now we have to find out the modulus of 2i−−2i\sqrt {2i} - \sqrt { - 2i}
⇒∣2i−−2i∣=∣±2∣=2\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.