Question

%29%29" id="TexFormula1" title="6/5 (cos(120) + isin(120)) 3/5 (cos(45) + i sin(45))" alt="6/5 (cos(120) + isin(120)) 3/5 (cos(45) + i sin(45))" align="absmiddle" class="latex-formula">

Answer 1

The product is equal to:

$\frac{18}{25}(cos(165) + i*sin(165))$

How to solve the product?

Remember that we can write a complex number in polar form as:

$R*e^{i*a} = R*(cos(a) + i*sin(a))$

Then the given product:

$\frac{6}{5}*(cos(120) + i*sin(120))*\frac{3}{5}*(cos(45) + i*sin(45))$

can be rewritten to:

$(\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})$

Now is easier to solve the product:

$(\frac{6}{5}*e^{i*120})*(\frac{3}{5}*e^{i*45})\\\\= \frac{6}{5} *\frac{3}{5} *e^{i*(120 + 45)}\\\\= \frac{18}{25}*e^{i*165}\\\\= \frac{18}{25}(cos(165) + i*sin(165))$

If you want to learn more about complex numbers:

brainly.com/question/10662770

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