<img src="https://tex.z-dn.net/?f=6%2F5%20%28cos%28120%29%20%2B%20isin%28120%29%29%203%2F5%20%28cos%2845%29%20%2B%20i%20sin%2845

2 years ago

%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">

Mathematics

1 answer:

marysya [2.9K]2 years ago

4 0

The product is equal to:

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

<h3>How to solve the product?</h3>

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:

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