Answer
$(2i) \cdot(3i) = (2\, \text{cis}\, 90^\circ)(3\, \text{cis}\, 90^\circ) = 6\,\text{cis}\, 180^\circ = -6$
Work Step by Step
$\underline{\text{Converting}\ 2i\ \text{and}\ 3i\ \text{to trigonometric form}}$
Note that $2i = 0 + 2i = x +iy$, where $x=0$ and $y=2$.
The modulus of $2i$ is $r=\sqrt{x^2+y^2} = \sqrt{0^2+2^2} = \sqrt{4} = 2$.
Its argument $\theta$ is such that $\sin\theta = y/r = 2/2 =1$ and $\cos\theta = x/r = 0/2 = 0$, so that we can put $\theta =90^\circ$.
Thus, $2i = 2\, \text{cis}\, 90^\circ$.
Similarly, $3i = 3\,\text{cis}\, 90^\circ$.
These are summarized in the table below.
$\small\begin{array}{|c|c|c|c|c|c|c|c|} \hline
z & x+iy & x & y & \text{modulus},\ r & \cos\theta = x/r & \sin\theta =y/r & \theta & r\,\text{cis}\,\theta \\ \hline
2i & 0 +2i & 0 & 2 & \sqrt{0^2+2^2} = \sqrt{4} = 2 & 0/2 = 0 & 2/2 = 1 & 90^\circ & 2\, \text{cis}\, 90^\circ \\ \hline
3i & 0 +3i & 0 &3 & \sqrt{0^2+3^2} = \sqrt{9} = 3 & 0/3 = 0 & 3/3 = 1 & 90^\circ & 3\, \text{cis}\, 90^\circ\\ \hline
\end{array}$
$\underline{\text{Product of}\ 2i, 3i} \\
\begin{align*}
(2i) \cdot(3i) &= (2\, \text{cis}\, 90^\circ)(3\, \text{cis}\, 90^\circ) \\
&= (2\cdot 3)\, \text{cis}\, (90^\circ +90^\circ) \\
&= 6\, \text{cis}\, 180^\circ\\
&= 6(\cos 180^\circ + i\sin 180^\circ) \\
&= 6(-1 + i(0)) \\
&= 6(-1) \\
(2i) \cdot(3i) &= -6
\end{align*}$
