Answer
$- \dfrac{15 \sqrt 2}{2}+i \dfrac{15 \sqrt 2}{2}$
Work Step by Step
The product theorem for two complex numbers is given by: $[r_1 (\cos \theta_1 +i \sin \theta_1) ] \cdot [r_2 (\cos \theta_2 +i \sin \theta_2) ]= r_1 r_2 [ \cos (\theta_1+\theta_2)+\sin (\theta_1+\theta_2)]$
This can also be re-written in the easy form as follows: $(r_1 \ cis \theta_1) \cdot ( r_2 \ cis \theta_2) = r_1 r_2 \ cis (\theta_1+\theta_2)$
Therefore, the given product becomes:
$(5 \ cis 90^{\circ}) \cdot ( 3 \ cis 45^{\circ}) =(5)(3) \ cis (90^{\circ} +45^{\circ})= 15 \ cis 135^{\circ}$
Re-arrange as: $15 \ cis 135^{\circ}= 15 (\cos 135^{\circ}+\ i \sin 135^{\circ}) $
Next, we will write it in the rectangular form by using calculator as follows:
$ 15 (\cos 135^{\circ}+\ i \sin 135^{\circ}) = 15 (-\dfrac{\sqrt 2}{2} +i \dfrac{\sqrt {2}} {2})= - \dfrac{15 \sqrt 2}{2}+i \dfrac{15 \sqrt 2}{2}$
