Question

Consider $i=\sqrt{-1}$ ,then the value of $\sqrt{i}+\sqrt{-i}$ is:
(a) 0
(b) $\sqrt{2}$
(c) -i
(d) i

Answer 1

Hint: Find the conditions given in question. As given terms are imaginary so try to convert them into the terms of Euler’s formula.
${{e}^{ix}}=\cos x+i\sin x$ . ${{e}^{ix}}$ can be written as $cisx$
$cisx=\cos x+i\sin x$

Complete step-by-step solution -
Definition of i, can be written as
The solution of the equation: ${{x}^{2}}+1=0$ is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as: An equation containing complex number in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0,x= -1$ is the root of the equation.
We’ve defined radical applied to real numbers as principal value. The value of complex square roots is a bit more complex, not that simple.
Definition of transcendental functions:
The functions which cannot be written in terms of algebraic functions are called transcendental functions.
The exponential formula is the immediate jump to polar coordinates. We take transcendental function of i then we can say the formula:
$\sqrt{a+bi}=\pm \left( \sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}+a}{2}}+i sgn \left( b \right)\sqrt{\dfrac{\sqrt{{{a}^{2}}+{{b}^{2}}}-a}{2}} \right)$
We know: sgn (0) = 0, sgn (positive numbers) = 1, sgn (negative numbers) = -1.
We have a = 0, b = 1
By substituting a, b values, we get:
$\sqrt{i}=\pm \left( \sqrt{\dfrac{1}{2}}+i\sqrt{\dfrac{1}{2}} \right)=\pm \dfrac{\left( 1+i \right)}{\sqrt{2}}$
By simplifying we get: $\sqrt{i}=\pm \dfrac{\left( 1+i \right)}{\sqrt{2}}$
Now we have, a = 0, b = -1
By substituting a, b into the equation, we get:
$\sqrt{-1}=\pm \dfrac{\left( 1-i \right)}{\sqrt{2}}$
By adding both equation we get 4 possibilities
$\sqrt{i}+\sqrt{-i}=\dfrac{\left( \pm \left( 1+i \right)\pm \left( 1-i \right) \right)}{\sqrt{2}}$
The brackets term has four cases
$\begin{align}
  & \left( 1+i \right)+\left( 1-i \right)=2 \\
 & \left( 1+i \right)-\left( 1-i \right)=2i \\
 & -\left( 1+i \right)+\left( 1-i \right)=-2i \\
 & -\left( 1+i \right)-\left( 1-i \right)=-2 \\
\end{align}$
The above equations are obtained by cancelling common terms
$\begin{align}
  & \sqrt{i}+\sqrt{-i}=\dfrac{2}{\sqrt{2}},\dfrac{2i}{\sqrt{2}},\dfrac{-2i}{\sqrt{2}},\dfrac{-2}{\sqrt{2}} \\
 & =\pm \sqrt{2},\pm \sqrt{2}i
\end{align}$
By options we have $\sqrt{2}$
Option (b) is true.

Note: Alternative is to use Euler’s formula.Here students need to be careful while converting in euler form of a complex number.
$\begin{align}
  & i=cis\dfrac{\pi }{2} \\
 & -i=cis\left( -\dfrac{\pi }{2} \right) \\
 & \sqrt{i}=cis\dfrac{\pi }{4} \\
 & \sqrt{-i}=cis\left( -\dfrac{\pi }{4} \right) \\
 & \sqrt{i}=\dfrac{\left( 1+i \right)}{\sqrt{2}} \\
 & \sqrt{-i}=\dfrac{\left( 1-i \right)}{\sqrt{2}} \\
 & \sqrt{i}+\sqrt{i}={{e}^{-i\dfrac{\pi }{4}}}+{{e}^{i\dfrac{\pi }{4}}} \\
 & \Rightarrow \sqrt{i}+\sqrt{-i}=\sqrt{2} \\
\end{align}$