How do you multiply # (-1+2i)(3-i) # in trigonometric form?

1 Answer
Apr 13, 2016

#(-1+2i)*(3-i)=sqrt50[cosarctan(-7)+isinarctan(-7)]#

Explanation:

We first covert them into trigonometric form. In this form #a+bi=r(costheta+isintheta)# or #a+bi=r*e^(itheta)#, where #r=sqrt(a^2+b^2)# and #theta=arctan(b/a)#

Hence, #-1+2i=sqrt5e^(ialpha)#, where #alpha=arctan(-2)#

and #3-i=sqrt10e^(ibeta)#, where #beta=arctan(-1/3)#

Hence #(-1+2i)*(3-i)=sqrt50e^(i(alpha+beta))#

As #tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalpha*tanbeta)# or

#tan(alpha+beta)=((-2)-1/3)/(1-(-2)*(-1/3))=(-7/3)/(1-2/3)=(-7/3)/(1/3)=-7#

Hence #(-1+2i)*(3-i)=sqrt50e^(i(arctan(-7)))#

or #(-1+2i)*(3-i)=sqrt50[cosarctan(-7)+isinarctan(-7)]#