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

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

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

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

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

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