Question

The value of \[\sqrt{i}+\sqrt{\left( -i \right)}\] is
(a) 0
(b) \[\sqrt{2}\]
(c) \[-i\]
(d) \[i\]

Answer 1

Hint: \[i\] and \[\left( -i \right)\] are conjugate. The sum will have 4 roots real and imaginary. \[-i={{90}^{\circ }}\], thus square root has \[{{45}^{\circ }}\] and \[\left( {{45}^{\circ }}+{{180}^{\circ }} \right)\]. Thus get the value of \[\sqrt{i}\] and \[\sqrt{-i}\], add them and find the value.

Complete step-by-step answer:
Here we have been given an expression, for which we need to find the value of \[i\] and \[\left( -i \right)\] are conjugates. Thus the sum of these conjugates will be real numbers.
But with complex numbers it is best to treat the non – integer powers and squares as multivated. So this particular expression can have 4 values. Two of these values of the sum of conjugate will be real numbers.
The complex number ‘\[i\]’ has an argument of \[{{90}^{\circ }}\]. Thus its has square roots at \[{{45}^{\circ }}\] and \[\left( {{45}^{\circ }}+{{180}^{\circ }} \right)\].
Now we know that a complex number, \[z=a+ib\].
\[\begin{align}
  & z=a+ib \\
 & z=\cos \theta +i\sin \theta \\
\end{align}\]
Hence put, \[\theta ={{45}^{\circ }}\].
From the trigonometric table we know that, \[\cos \theta =\sin \theta =\dfrac{1}{\sqrt{2}}\].
\[\begin{align}
  & z=\cos \theta +i\sin \theta \\
 & z=\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}\left( 1+i \right) \\
\end{align}\]
Hence, \[\sqrt{i}=\pm \dfrac{1}{\sqrt{2}}\left( 1+i \right)\]
Similarly for, \[\sqrt{-i}=\pm \dfrac{1}{\sqrt{2}}\left( 1-i \right)\].
Hence, \[\sqrt{i}+\sqrt{-i}=\pm \dfrac{1}{\sqrt{2}}\left( 1+i \right)\pm \dfrac{1}{\sqrt{2}}\left( 1-i \right)\]
Now let us find the 4 values by considering different cases,
Case 1: \[\sqrt{i}+\sqrt{-i}=\dfrac{1}{\sqrt{2}}\left( 1+i \right)+\dfrac{1}{\sqrt{2}}\left( 1-i \right)\]
\[\sqrt{i}+\sqrt{-i}=\dfrac{1}{\sqrt{2}}\left( 1+i+1-i \right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\]
Case 2: \[\sqrt{i}+\sqrt{-i}=\dfrac{1}{\sqrt{2}}\left( 1+i \right)-\dfrac{1}{\sqrt{2}}\left( 1-i \right)\]
\[\sqrt{i}+\sqrt{-i}=\dfrac{1}{\sqrt{2}}\left( 1+i-1+i \right)=\dfrac{2i}{\sqrt{2}}=\sqrt{2}i\]
Case 3: \[\sqrt{i}+\sqrt{-i}=\dfrac{-1}{\sqrt{2}}\left( 1+i \right)+\dfrac{1}{\sqrt{2}}\left( 1-i \right)\]
\[\sqrt{i}+\sqrt{-i}=\dfrac{1}{\sqrt{2}}\left[ -1-i+1-i \right]=\dfrac{-2i}{\sqrt{2}}=-\sqrt{2}i\]
Case 4: \[\sqrt{i}+\sqrt{-i}=\dfrac{-1}{\sqrt{2}}\left( 1+i \right)-\dfrac{1}{\sqrt{2}}\left( 1-i \right)\]
\[\sqrt{i}+\sqrt{-i}=\dfrac{1}{\sqrt{2}}\left[ -1-i-1+i \right]=\dfrac{-2}{\sqrt{2}}=-\sqrt{2}\]
Hence, we got the four values \[\sqrt{2}\], \[-\sqrt{2}\], \[\sqrt{2}i\], \[-\sqrt{2}i\]. Out of which \[\sqrt{2}\] and \[-\sqrt{2}\] are real numbers.
Thus comparing without given option we get \[\sqrt{2}\] as the value.
Hence, \[\sqrt{i}+\sqrt{-i}=\sqrt{2}\].
\[\therefore \] Option (b) is the correct answer.

Note: You can also find the value by squaring the expression and take its square root. i.e. \[\sqrt{i}+\sqrt{-i}\]
\[\begin{align}
  & \sqrt{i}+\sqrt{-i}=\sqrt{{{\left( \sqrt{i}+\sqrt{-i} \right)}^{2}}}\left\{ \because {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}} \right\} \\
 & \sqrt{i}+\sqrt{-i}=\sqrt{{{\left( \sqrt{i} \right)}^{2}}+2\left( \sqrt{i}\sqrt{-i} \right)+{{\left( \sqrt{-i} \right)}^{2}}} \\
 & \sqrt{i}+\sqrt{-i}=\sqrt{i+2\sqrt{i\times \left( -i \right)}+\left( -i \right)}=\sqrt{i-i+2\sqrt{-{{i}^{2}}}}\left\{ \because {{i}^{2}}=\left( -1 \right)\Rightarrow \sqrt{1}=1 \right\} \\
 & \sqrt{i}+\sqrt{-i}=\sqrt{0+2\sqrt{-\left( -1 \right)}}=\sqrt{2\sqrt{1}} \\
 & \sqrt{i}+\sqrt{-i}=\sqrt{2} \\
\end{align}\]
Thus we got, \[\sqrt{i}+\sqrt{-i}=\sqrt{2}\].