Answer
The two square roots are:
\[
-\sqrt{2}-i \sqrt{2} \text { and }\sqrt{2}+i \sqrt{2}
\]
Work Step by Step
\[
0+4 i=4 i
\]
The modulus of a complex number in standard form $y+x$ is:
\[
\begin{array}{c}
\sqrt{y^{2}+x^{2}}=r \\
\sqrt{4^{2}+0^{2}}=r \\
\sqrt{16}=r \\
4=r
\end{array}
\]
Angle $\theta$ is the smallest positive angle for which:
\[
0=\frac{0}{4}=\frac{x}{r}=\cos \theta
\]
,
\[
1=\frac{4}{4}=\frac{y}{r}=\sin \theta
\]
Thus
\[
90^{\circ}=\theta
\]
Using the definition of the trigonometric form of a complex number:
\[
\begin{aligned}
&=r(\cos \theta+i \sin \theta) \\
&=4\left(\cos 90^{\circ}+i \sin 90^{\circ}\right)
\end{aligned}
\]
Using the roots theorem:
\[
u_{k}=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]
\]
$\mathrm{W} \operatorname{hen} 0=k$
\[
\begin{array}{c}
\mathrm{u}_{\mathrm{u}}=4^{\mathrm{t}}\left[\cos \left(\frac{90^{\circ}}{2}+\frac{360^{\circ}}{2}(0)\right)+i \sin \left(\frac{90^{\circ}}{2}+\frac{36 \mathrm{fr}}{2}(0)\right)\right] \\
=2 \cos 45^{\circ}+i \sin 45^{\circ} \\
=2\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\right]
\end{array}
\]
Multiply:
\[
=\sqrt{2}+i \sqrt{2}
\]
When $1=k$
\[
\begin{array}{c}
\mathrm{u}_{1}=4 \pm\left[\cos \left(\frac{90^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)+i \sin \left(\frac{90^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)\right] \\
=2\left[\cos \left(45^{\circ}+180^{\circ}\right)+i \sin \left(45^{\circ}+180^{\circ}\right)\right] \\
\left.=2 \cos 225^{\circ}+i \sin 225^{\circ}\right] \\
=\left[-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i\right]2
\end{array}
\]
Multiply:
\[
=[-\sqrt{2}-t \sqrt{2}]
\]
