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

1 Answer
Apr 3, 2018

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

Explanation:

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)])#