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, $z=a+ib$.

& z=a+ib \\\ & z=\cos \theta +i\sin \theta \\\ \end{aligned}$$ Hence put, $$\theta ={{45}^{\circ }}$$. From the trigonometric table we know that, $$\cos \theta =\sin \theta =\dfrac{1}{\sqrt{2}}$$. $$\begin{aligned} & 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{aligned}$$ Hence, $$\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{aligned} & \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{aligned}$$ Thus we got, $$\sqrt{i}+\sqrt{-i}=\sqrt{2}$$.