Question

The real and imaginary parts of $\sqrt{i}$ are, where $i=\sqrt{-1}$
[a] $1-\dfrac{1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}}$ respectively
[b] $1-\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}$ respectively
[c] $1+\dfrac{1}{\sqrt{2}},\dfrac{1}{\sqrt{2}}$ respectively
[d] $1+\dfrac{1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}}$ respectively

Answer 1

Hint: Assume that $\sqrt{i}=a+ib$. Square both sides and use the fact that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$. Compare the real and imaginary parts and hence form two equations in two variables a and b. Solve the equations and hence find the value of a and b. Hence find the value of the real and imaginary part of $1+\sqrt{i}$. Alternatively, use the fact that $i={{e}^{i\dfrac{\pi }{2}}}={{e}^{i\dfrac{5\pi }{2}}}$. Use the fact that ${{e}^{i\theta }}=\cos \theta +i\sin \theta$ and hence find the real and imaginary parts of $1+\sqrt{i}$.

Complete step-by-step solution
Let $\sqrt{i}=a+ib,a,b\in \mathbb{R}$
Squaring both sides, we get
$i={{\left( a+ib \right)}^{2}}$
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Hence, we have
$i={{a}^{2}}+{{\left( ib \right)}^{2}}+2a\left( ib \right)$
We know that ${{i}^{2}}=-1$. Hence, we get
$i={{a}^{2}}-{{b}^{2}}+i2ab$
Comparing real parts, we get
$\begin{aligned} & {{a}^{2}}-{{b}^{2}}=0 \\\ & \Rightarrow a=\pm b\text{ }\left( i \right) \\\ \end{aligned}$
Comparing imaginary parts, we get
$2ab=1$
Substituting the value of a from equation (i), we get
$\begin{aligned} & \pm 2{{b}^{2}}=1 \\\ & \Rightarrow {{b}^{2}}=\pm \dfrac{1}{2} \\\ \end{aligned}$
Since $b\in \mathbb{R}$, we have ${{b}^{2}}\ge 0$
Hence we have
${{b}^{2}}=-\dfrac{1}{2}$ is rejected.
Hence, we have
$\begin{aligned} & {{b}^{2}}=\dfrac{1}{2} \\\ & \Rightarrow b=\pm \dfrac{1}{\sqrt{2}} \\\ \end{aligned}$
Hence, from equation (i), we get
$a=\pm \dfrac{1}{\sqrt{2}}$
Hence, we have
$\sqrt{i}=\pm \dfrac{1}{\sqrt{2}}\pm i\dfrac{1}{\sqrt{2}}=\pm \dfrac{1}{\sqrt{2}}\left( 1+i \right)$
Hence, we have
$1+\sqrt{i}=1\pm \dfrac{1}{\sqrt{2}}+i\left( \pm \dfrac{1}{\sqrt{2}} \right)$
Hence, we have $\operatorname{Re}\left( 1+\sqrt{i} \right)=1+\dfrac{1}{\sqrt{2}},\operatorname{Im}\left( 1+\sqrt{i} \right)=\dfrac{1}{\sqrt{2}}\text{ or }\operatorname{Re}\left( 1+\sqrt{i} \right)=1-\dfrac{1}{\sqrt{2}},\operatorname{Im}\left( 1+\sqrt{i} \right)=-\dfrac{1}{\sqrt{2}}$
Hence options [a] and [c] are correct.

Note: Alternative solution: Using Euler’s identity:
We know that ${{e}^{i\theta }}=\cos \theta +i\sin \theta$
Hence, we have
$\begin{aligned} & {{e}^{i\dfrac{\pi }{2}}}=\cos \dfrac{\pi }{2}+i\sin \dfrac{\pi }{2}=i \\\ & {{e}^{i\dfrac{5\pi }{2}}}=\cos \dfrac{5\pi }{2}+i\sin \dfrac{5\pi }{2}=i \\\ \end{aligned}$
Hence, we have
$\sqrt{i}={{\left( {{e}^{\dfrac{i\pi }{2}}} \right)}^{\dfrac{1}{2}}}={{e}^{i\dfrac{\pi }{4}}}=\cos \dfrac{\pi }{4}+i\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}$
Also, we have
$\sqrt{i}={{\left( {{e}^{i\dfrac{5\pi }{2}}} \right)}^{\dfrac{1}{2}}}={{e}^{\dfrac{5\pi }{4}}}=\cos \dfrac{5\pi }{4}+i\sin \dfrac{5\pi }{4}=-\dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}}$
Hence, we have
$\sqrt{i}=\pm \dfrac{1}{\sqrt{2}}\pm \dfrac{i}{\sqrt{2}}$, which is the same as obtained above.
Hence following a similar procedure as above, we get
$\operatorname{Re}\left( 1+\sqrt{i} \right)=1+\dfrac{1}{\sqrt{2}},\operatorname{Im}\left( 1+\sqrt{i} \right)=\dfrac{1}{\sqrt{2}}\text{ or }\operatorname{Re}\left( 1+\sqrt{i} \right)=1-\dfrac{1}{\sqrt{2}},\operatorname{Im}\left( 1+\sqrt{i} \right)=-\dfrac{1}{\sqrt{2}}$
Hence options [a] and [c] are correct.