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

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

Explanation

Solution

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

Complete step-by-step solution -
Definition of i, can be written as
The solution of the equation: x2+1=0{{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: (1+i)x+(1+i)=0,x=−1\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:
a+bi=±(a2+b2+a2+isgn(b)a2+b2−a2)\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:
i=±(12+i12)=±(1+i)2\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: i=±(1+i)2\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:
−1=±(1−i)2\sqrt{-1}=\pm \dfrac{\left( 1-i \right)}{\sqrt{2}}
By adding both equation we get 4 possibilities
i+−i=(±(1+i)±(1−i))2\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
(1+i)+(1−i)=2 (1+i)−(1−i)=2i −(1+i)+(1−i)=−2i −(1+i)−(1−i)=−2 \begin{aligned} & \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{aligned}
The above equations are obtained by cancelling common terms
i+−i=22,2i2,−2i2,−22 =±2,±2i\begin{aligned} & \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{aligned}
By options we have 2\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.
i=cisπ2 −i=cis(−π2) i=cisπ4 −i=cis(−π4) i=(1+i)2 −i=(1−i)2 i+i=e−iπ4+eiπ4 ⇒i+−i=2 \begin{aligned} & 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{aligned}