The product is equal to:
[tex]\frac{18}{25}(cos(165) + i*sin(165))[/tex]
Remember that we can write a complex number in polar form as:
[tex]R*e^{i*a} = R*(cos(a) + i*sin(a))[/tex]
Then the given product:
[tex]\frac{6}{5}*(cos(120) + i*sin(120))*\frac{3}{5}*(cos(45) + i*sin(45))[/tex]
can be rewritten to:
[tex](\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})[/tex]
Now is easier to solve the product:
[tex](\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})\\\\= \frac{6}{5} *\frac{3}{5} *e^{i*(120 + 45)}\\\\= \frac{18}{25}*e^{i*165}\\\\= \frac{18}{25}(cos(165) + i*sin(165))[/tex]
If you want to learn more about complex numbers:
https://brainly.com/question/10662770
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