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

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

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Answer 1 · 2016-05-19T09:34:18

$(2 - i) \cdot (1 - 2 i) = - 5 i$ $(2-i)*(1-2i)=-5i$

Explanation:

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

$a + i b$ $a+ib$ can be written in trigonometric form $r e^{i θ} = r cos θ + i r sin θ = r (cos θ + i sin θ)$ $re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)$ ,
where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ .

Hence $2 - i = \sqrt{2^{2} + {(- 1)}^{2}} [cos α + i sin α]$ $2-i=sqrt(2^2+(-1)^2)[cosalpha+isinalpha]$ or

$\sqrt{5} e^{i α}$ $sqrt5e^(ialpha)$ , where $cos α = \frac{2}{\sqrt{5}}$ $cosalpha=2/sqrt5$ and $sin α = - \frac{1}{\sqrt{5}}$ $sinalpha=-1/sqrt5$

$1 - 2 i = \sqrt{1^{2} + {(- 2)}^{2}} [cos β + i sin β]$ $1-2i=sqrt(1^2+(-2)^2)[cosbeta+isinbeta]$ or

$\sqrt{5} e^{i β}$ $sqrt5e^(ibeta]$ , where $cos β = \frac{1}{\sqrt{5}}$ $cosbeta=1/sqrt5$ and $sin β = \frac{- 2}{\sqrt{5}}$ $sinbeta=(-2)/sqrt5$

Hence $(2 - i) \cdot (1 - 2 i) = (\sqrt{5} e^{i α}) \times (\sqrt{5} e^{i β}) = 5 e^{i (α + β)}$ $(2-i)*(1-2i)=(sqrt5e^(ialpha))xx(sqrt5e^(ibeta])=5e^(i(alpha+beta))$

= $5 (cos (α + β) + i sin (α + β))$ $5(cos(alpha+beta)+isin(alpha+beta))$

= $5 [(cos α cos β - sin α sin β) + i (sin α cos β + cos α sin β)]$ $5[(cosalphacosbeta-sinalphasinbeta)+i(sinalphacosbeta+cosalphasinbeta)]$

= $5 [(\frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} - \frac{- 1}{\sqrt{5}} \times \frac{- 2}{\sqrt{5}}) + i (\frac{- 1}{\sqrt{5}} \times \frac{1}{\sqrt{5}} + \frac{2}{\sqrt{5}} \times \frac{- 2}{\sqrt{5}})]$ $5[(2/sqrt5xx1/sqrt5-(-1)/sqrt5xx(-2)/sqrt5)+i((-1)/sqrt5xx1/sqrt5+2/sqrt5xx(-2)/sqrt5)]$

= $5 [(\frac{2}{5} - \frac{2}{5}) + i (- \frac{1}{5} - \frac{4}{5})]$ $5[(2/5-2/5)+i(-1/5-4/5)]$ = $5 \times - \frac{5}{5} i = - 5 i$ $5xx-5/5i=-5i$

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

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