How do you multiply $(4-5i)(2+7i)$ in trigonometric form?

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Answer 1 · 2018-04-03T03:36:09

$color(purple)(z_1z_2=sqrt2173[cos(0.396)+isin(0.396)])$

Convert the numbers to polar form.

$color(red)(z_1=4-5i)$

$rArrr_1=sqrt(4^2+(-5)^2)=sqrt(16+25)=sqrt41$

$rArrtheta_1=arctan((-5)/4)~~5.387$

$rArrr_1[costheta_1+isintheta_1]=color(red)(sqrt41[cos(5.387)+isin(5.387)])$

$color(blue)(z_2=2+7i)$

$rArrr_2=sqrt(2^2+7^2)=sqrt(4+49)=sqrt53$

$rArrtheta_2=arctan(7/2)~~1.292$

$rArrr_2[costheta_2+isintheta_2]=color(blue)(sqrt53[cos(1.292)+isin(1.292)])$

Now to multiply them together, the result will be:

$z_1z_2=r_1r_2[cos(theta_1+theta_2)+isin(theta_1+theta_2)]$

$rArrr_1r_2=sqrt41sqrt53=sqrt(41*53)=sqrt(2173)$

$rArrtheta_1+theta_2=arctan((-5)/4)+arctan(7/2)~~6.680$

We usually try to express $theta$ on the interval

$0 < theta < 2pi$

$6.680 - 2pi~~0.396$

So our final answer is:

$color(purple)(z_1z_2=sqrt2173[cos(0.396)+isin(0.396)])$

How do you multiply (4-5i)(2+7i) (4-5i)(2+7i) in trigonometric form?