Root of iota?

Using $\color{#3D99F6}{i^2+1=0}$ $\large \sqrt{-i}=\frac{1}{\sqrt2}(\sqrt{-2i})$ $\large =\frac{1}{\sqrt2}(\sqrt{\color{#3D99F6}{i^2+1}-2i})$ $\large =\frac{1}{\sqrt2}\sqrt{(i-1)^2}$ $\Large=\pm \frac { 1 }{ \sqrt { 2 } } \mp \frac { 1 }{ \sqrt { 2 } } i$

$let\quad the\quad roots\quad be\quad x+y\iota \quad i.e\sqrt { -\iota } \\ ({ x+y\iota ) }^{ 2 }=-\iota \\ { x }^{ 2 }-{ y }^{ 2 }+2xy\iota =-\iota \\ comparing\quad real\quad and\quad imaginary\quad parts,\\ { x }^{ 2 }-{ y }^{ 2 }=0\quad eq(1)\\ 2xy=-\iota \quad eq(2)\\ using\quad identity,\\ { { (x }^{ 2 }+{ y }^{ 2 }) }^{ 2 }=({ { x }^{ 2 }-{ y }^{ 2 }) }^{ 2 }+({ 2xy) }^{ 2 }\\ \therefore \quad { x }^{ 2 }+{ y }^{ 2 }=1\quad eq(3)\quad \quad \quad \quad \quad \quad (As\quad square\quad of\quad real\quad no.\quad cannot\quad be\quad negative)\\ from\quad eq(1)\quad \& \quad eq(3)\\ x=\pm \frac { 1 }{ \sqrt { 2 } } \\ similarly,\\ y=\pm \frac { 1 }{ \sqrt { 2 } } \\ \therefore \quad The\quad roots\quad are\quad \pm \frac { 1 }{ \sqrt { 2 } } \mp \frac { 1 }{ \sqrt { 2 } } \iota \\ (because\quad both\quad x\quad and\quad y\quad can't\quad be\quad \quad simultaneously \quad positive\quad or\quad negative\quad as\quad their\quad product\quad is\quad negative\quad eq(2).$