One value of ${\left(1+\mathrm{i}\right)}^{1/2}\mathrm{is} 2^{1/4}{\mathrm{e}}^{\mathrm{i\pi }/8}.$

The value is

$\begin{array}{rcl}{(1+\mathrm{i})}^{1/2}& =& {\sqrt{2}}^{1/2} {\left(\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}\right)}^{1/2}\\ & =& 2^{1/4}{\left(\cos \frac{\mathrm{\pi }}{4}+\mathrm{i} \sin \frac{\mathrm{\pi }}{4}\right)}^{1/2}\\ & =& 2^{1/4} \left(\cos \frac{2\mathrm{k\pi }+{\displaystyle \frac{\mathrm{\pi }}{4}}}{2}+\mathrm{i} \sin \frac{2\mathrm{k\pi }+{\displaystyle \frac{\mathrm{\pi }}{4}}}{2}\right) , \mathrm{k}=0, 1\\ & =& 2^{1/4} \mathrm{cis}\frac{\mathrm{\pi }}{8} , 2^{1/4} \mathrm{cis} \frac{9\mathrm{\pi }}{8} \end{array}$

one given root is $2^{1/4} {\mathrm{e}}^{\mathrm{i\pi }/8}$

$\therefore$ other root is $2^{1/4} {\mathrm{e}}^{\mathrm{i}\left(\raisebox{1ex}{$9\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)}$