Given: Two complex numbers below
[tex]\begin{gathered} z_1=2+2i \\ z_2=-3+3i \end{gathered}[/tex]
To Determine: The product of the given complex numbers
[tex]z_1z_2=(2+2i)(-3+3i)[/tex][tex]\begin{gathered} z_1z_2=2(-3+3i)+2i(-3+3i) \\ z_1z_2=-6+6i-6i+6i^2 \\ z_1z_2=-6+6i^2 \end{gathered}[/tex]
Please note that
[tex]\begin{gathered} i=\sqrt[]{-1} \\ i^2=(\sqrt[]{-1})^2_{} \\ i^2=-1 \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} z_1z_2=-6+6i \\ z_1z_2=-6+6(-1) \\ z_1z_2=-6-6 \\ z_1z_2=-12+0i \end{gathered}[/tex]
Let us convert the product to polar form
Please note that
[tex]\begin{gathered} if,z=x+iy,the\text{ polar form is} \\ z=r(\cos \theta+i\sin \theta) \\ \text{where} \\ r=\sqrt[]{x^2+y^2} \\ \tan \theta=\frac{y}{x} \\ \theta=tan^{-1}(\frac{y}{x}) \end{gathered}[/tex]
Apply the conversion into the product we got
[tex]\begin{gathered} z_1z_2=-12+0i,x=-12,y=0 \\ r=\sqrt[]{x^2+y^2}=\sqrt[]{(-12)^2+0^2} \\ r=\sqrt[]{144+0} \\ r=\sqrt[]{144} \\ r=12 \\ \theta=\tan ^{-1}(\frac{0}{-12}) \\ \theta=\tan ^{-1}(0) \\ \theta=\pi \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} z_1z_2=r(\cos \theta+i\sin \theta) \\ r=12,\theta=\pi \\ z_1z_2=12(\cos \pi+i\sin \pi) \end{gathered}[/tex]
Hence, the product of the complex numbers in polar form is
12(cosπ+isinπ)
