Answer
Choice a
\[
3 \sqrt{2}+3 i \sqrt{2} \ \ \ \ \text {and} \quad-3 \sqrt{2}-3 i \sqrt{2}
\]
Work Step by Step
The modulus of a complex number in the standard form $x+y i$ is:
\[
36 i+0=36 i
\]
\[
\begin{array}{l}
=\sqrt{y^{2}+x^{2}} \\
=\sqrt{36^{2}+0^{2}} \\
=\sqrt{1296} \\
=36
\end{array}
\]
Angle $\theta$ is the smallest positive angle for which:
\[
1=\frac{36}{36}=\frac{x}{r}=\cos \theta
\]
\[
0=\frac{0}{36}=\frac{y}{r}=\sin \theta
\]
$\mathrm{Thus}$
\[
90^{\circ}=\theta
\]
Using the definition of the trigonometric form of a complex number:
\[
\begin{aligned}
&=(\cos \theta+i \sin \theta) r\\
&36\left(\cos 90^{\circ}+i \sin 90^{\circ}\right)=36 i
\end{aligned}
\]
Using the roots theorem:
\[
r^{\frac{1}{n}}\left[\cos \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)+i \sin \left(\frac{\theta}{n}+\frac{360^{\circ}}{n} k\right)\right]=u_{k}
\]
When $0=k$ :
\[
36^{\frac{1}{2}}\left[\cos \left(\frac{90^{\circ}}{2}+\frac{360^{\circ}}{2}(0)\right)+i \sin \left(\frac{90^{\circ}}{2}+\frac{360^{\circ}}{2}(0)\right)\right]=u_{0}
\]
Simplify:
\[
\begin{aligned}
=& [\cos 45+i \sin 45] 6\\
&=\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} i\right]6
\end{aligned}
\]
Multiply:
\[
\begin{array}{l}
=\frac{6}{\sqrt{2}}+\frac{6}{\sqrt{2}} i \\
=[3 \sqrt{2}+3 i \sqrt{2}]
\end{array}
\]
When $1=k$ :
\[
36^{\frac{1}{2}}\left[\cos \left(\frac{90^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)+i \sin \left(\frac{90^{\circ}}{2}+\frac{360^{\circ}}{2}(1)\right)\right]=w_{1}
\]
Simplify:
\[
\begin{array}{c}
=\left[\cos \left(45^{\circ}+180^{\circ}\right)+i \sin \left(45^{\circ}+180^{\circ}\right)\right]6 \\
=\left[\cos 225^{\circ}+i \sin 225^{\circ}\right] 6\\
=\left[-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} i\right]6
\end{array}
\]
Multiply:
\[
\begin{array}{l}
=-\frac{6}{\sqrt{2}}-\frac{6}{\sqrt{2}} i \\
=-3 \sqrt{2}-3 i \sqrt{2}
\end{array}
\]
Thus, the correct choice is a.
