Answer
$(4i) \cdot(2) = (4\, \text{cis}\, 90^\circ)(2\, \text{cis}\, 0^\circ) = 8\,\text{cis}\, 90^\circ = 8i$
Work Step by Step
$\underline{\text{Converting}\ 4i\ \text{and}\ 2\ \text{to trigonometric form}}$
First, write $4i = 0 + 4i = x +iy$, where $x=0$ and $y=4$.
The modulus of $4i$ is $r=\sqrt{x^2+y^2} = \sqrt{0^2+4^2} = \sqrt{16} = 4$.
The argument $\theta$ of $4i$ is such that $\sin\theta = y/r = 4/4 =1$ and $\cos\theta = x/r = 0/4 = 0$, so we can choose $\theta =90^\circ$.
Thus, $4i = 4\, \text{cis}\, 90^\circ$.
Note, $2 = 2 +0i$, where $x=2$ and $y=0$. The modulus is $r=\sqrt{2^2+0^2} = \sqrt{4}=2$, and the argument can be set to $\theta =0^\circ$ as $\cos\theta=x/r = 2/2=1$ and $\sin\theta=y/r=0/2=0$.
Thus, $2 = 2\,\text{cis}\, 0^\circ$.
These are summarized in the table below.
$\scriptsize\begin{array}{|c|c|c|c|c|c|c|c|} \hline
z & x+iy & x & y & \text{modulus},\ r & \cos\theta = x/r & \sin\theta =y/r & \theta & r\,\text{cis}\,\theta \\ \hline
4i & 0 +4i & 0 & 4 & \sqrt{0^2+4^2} = \sqrt{16} = 4 & 0/4 = 0 & 4/4 = 1 & 90^\circ & 4\, \text{cis}\, 90^\circ \\ \hline
2 & 2 +0i & 2 &0 & \sqrt{2^2+0^2} = \sqrt{4} = 2 & 2/2 = 1 & 0/2 = 0 & 0^\circ & 2\, \text{cis}\, 0^\circ\\ \hline
\end{array}$
$\underline{\text{Product of}\ 4i, 2} \\
\begin{align*}
(4i) \cdot(2) &= (4\, \text{cis}\, 90^\circ)(2\, \text{cis}\, 0^\circ) \\
&= (4\cdot 2)\, \text{cis}\, (90^\circ + 0^\circ) \\
&= 8\, \text{cis}\, 90^\circ\\
&= 8(\cos 90^\circ + i\sin 90^\circ) \\
&= 8(0 + i(1)) \\
&= 8(i) \\
(4i) \cdot(2) &= 8i
\end{align*}$
