What is the trigonometric form of # (2-i)*(1-2i) #?

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Answer 1 · 2016-05-19T09:34:18

#(2-i)*(1-2i)=-5i#

Let us first write #(2-i)# and #(1-2i)# in trigonometric form.

#a+ib# can be written in trigonometric form #re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)#,
where #r=sqrt(a^2+b^2)#.

Hence #2-i=sqrt(2^2+(-1)^2)[cosalpha+isinalpha]# or

#sqrt5e^(ialpha)#, where #cosalpha=2/sqrt5# and #sinalpha=-1/sqrt5#

#1-2i=sqrt(1^2+(-2)^2)[cosbeta+isinbeta]# or

#sqrt5e^(ibeta]#, where #cosbeta=1/sqrt5# and #sinbeta=(-2)/sqrt5#

Hence #(2-i)*(1-2i)=(sqrt5e^(ialpha))xx(sqrt5e^(ibeta])=5e^(i(alpha+beta))#

= #5(cos(alpha+beta)+isin(alpha+beta))#

= #5[(cosalphacosbeta-sinalphasinbeta)+i(sinalphacosbeta+cosalphasinbeta)]#

= #5[(2/sqrt5xx1/sqrt5-(-1)/sqrt5xx(-2)/sqrt5)+i((-1)/sqrt5xx1/sqrt5+2/sqrt5xx(-2)/sqrt5)]#

= #5[(2/5-2/5)+i(-1/5-4/5)]# = #5xx-5/5i=-5i#