Answer
$-\frac{15}{\sqrt{2}}+\frac{15}{\sqrt{2}}i$
Work Step by Step
First, we use the product theorem to multiply the absolute values and add the arguments:
$(5 cis 90^{\circ})(3 cis (45^{\circ}))
\\=5(3) cis (90^{\circ}+45^{\circ})
\\=15 cis (135^{\circ})$
Next, we change the expression into its equivalent form:
$=15 cis (135^{\circ})
\\=15(\cos135^{\circ}+i\sin135^{\circ})$
Since we know that $\cos135^{\circ}=-\frac{1}{\sqrt{2}}$ and $\sin135^{\circ}=\frac{1}{\sqrt{2}}$, we substitute these values in the expression and simplify:
$15(\cos90^{\circ}+i\sin90^{\circ})
\\=15(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)
\\=-\frac{15}{\sqrt{2}}+\frac{15}{\sqrt{2}}i$
