Answer
The two square roots are:
$5 i$ and $-5 i$
Work Step by Step
\[
-25+0 i=-25
\]
The modulus of a complex number in the standard form $x+y i$ is:
\[
\begin{array}{c}
\sqrt{x^{2}+y^{2}} =r\\
\sqrt{(-25)^{2}+0^{2}}=r \\
\sqrt{625}=r \\
25
=r
\end{array}
\]
Angle $\theta$ is the smallest positive angle for which:
\[
-1=\frac{x}{r}=\frac{-25}{25}=\cos \theta
\]
,
\[
\sin \theta=\frac{y}{r}=\frac{0}{25}=0
\]
Thus
\[
180^{\circ}=\theta
\]
Using the definition of the trigonometric form of a complex number:
\[
\begin{aligned}
&=(\cos \theta+i \sin \theta)r \\
&=25\left(\cos 180^{\circ}+i \sin 180^{\circ}\right)
\end{aligned}
\]
Using the roots theorem:
\[
r^{\frac{1}{n}}\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]=w_{k}
\]
When $0=k$ :
\[
\begin{array}{c}
\mathrm{u}_{0}=25^{\frac{1}{2}}\left[\cos \left(\frac{180^{\circ}}{2}+\frac{360^{\circ}}{2}(0)\right)+i \sin \left(\frac{180^{\circ}}{2}+\frac{360^{\circ}}{2}(0)\right)\right] \\
=5\left[\cos 90^{\circ}+i \sin 90^{\circ}\right] \\
=5[0+i]
\end{array}
\]
Multiply:
\[
=\sqrt{5 i}
\]
When $1=k$ :
\[
\begin{aligned}
25 !\left[\cos \left(\frac{180^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)+i \sin \left(\frac{180^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)\right]=u_{1} \\
=5\left[\cos \left(90^{\circ}+180^{\circ}\right)+i \sin \left(90^{\circ}+180^{\circ}\right)\right] & \\
=5\left[\cos 270^{\circ}+i \sin 270^{\circ}\right] &=[0-i]5
\end{aligned}
\]
= $-5 i$
