Answer
See explanation.
Work Step by Step
\begin{array}{l}
2=25=\left(60 \times 0^{\circ}+17 \sin 0^{\circ}\right) 25\\
\text { We know, } \left(\cos 360^{\circ}+15 \sin 360^{\circ}\right) 25\\
\sqrt{12}\left(\cos \frac{0}{2}+i \sin \frac{\theta}{2}\right)=z^{1 / 2} \\
=\left(\cos \frac{0^{\circ}}{2}+\hat{i} \sin \frac{0^{\circ}}{2}\right)25^{1 / 2}
\end{array}
\begin{array}{l}
=\left(\cos \frac{360^{\circ}}{2}+i \sin \frac{360^{\circ}}{2}\right)25^{1 / 2} \\
=\left(\cos 0^{\circ}+\pi \sin 0^{\circ}\right)5 \\
\left(\cos 180^{\circ}+i \sin 180^{\circ}\right)5 \\
=5(1) \operatorname{arcs}(-1) \\
=50 \alpha-5 \\
-5 \text { or }5=z^{\prime \prime}
\end{array}
