Answer
$2i$
Work Step by Step
First, we use the division theorem to divide the absolute values and subtract the arguments:
$\frac{12(\cos23^{\circ}+i\sin23^{\circ})}{6(\cos293^{\circ}+i\sin 293^{\circ})}
\\=2(\cos (23^{\circ}-293^{\circ})+i\sin(23^{\circ}-293^{\circ}))
\\=2(\cos-270^{\circ}+i\sin-270^{\circ})$
Since we know that $\cos-270^{\circ}=\cos270^{\circ}=0$ and $\sin-270^{\circ}=-\sin270^{\circ}=1$, we can substitute these values in the expression and simplify:
$2(\cos-270^{\circ}+i\sin-270^{\circ})
\\=2(i)
\\=2i$
