How do you divide $\frac{1 - 2 i}{6 + i}$ $(1-2i) / (6+i)$ in trigonometric form?

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Answer 1 · 2018-01-30T12:42:16

In trigonometric form: $0.368 (cos 5.011 + i sin 5.011)$ $0.368(cos 5.011+isin 5.011)$

$Z = \frac{1 - 2 i}{6 + i}$ $Z=(1-2i)/(6+i)$

$Z = a + i b$ $Z=a+ib$ . Modulus: $| Z | = \sqrt{a^{2} + b^{2}}$ $|Z|=sqrt (a^2+b^2)$ ;

Argument: $θ = {tan}^{- 1} (\frac{b}{a})$ $theta=tan^-1(b/a)$ Trigonometrical form :

$Z = | Z | (cos θ + i sin θ)$ $Z =|Z|(costheta+isintheta)$

$Z_{1} = 1 - 2 i$ $Z_1= 1-2 i$ .Modulus: $| Z | = \sqrt{1^{2} + {(- 2)}^{2}}$ $|Z|=sqrt(1^2+(-2)^2)$

$= \sqrt{5} \approx 2.236$ $=sqrt 5 ~~ 2.236$ Argument: $tan α = \frac{| - 2 |}{| 1 |}$ $tan alpha= (|-2|)/(|1|)$

$= 2$ $=2$ . $α = {tan}^{- 1} 2 = 1.107; Z_{1}$ $alpha =tan^-1 2 = 1.107; Z_1$ lies on fourth quadrant, so

$θ = 2 π - α = 2 π - 1.107 \approx 5.176$ $theta =2pi-alpha=2pi-1.107 ~~ 5.176$

$:. Z_1=2.236(cos5.176+isin 5.176)$ ,

$Z_2= 6 + i$ .Modulus: $|Z|=sqrt(6^2+1^2)$

$=sqrt 37 ~~ 6.083$ Argument: $tan alpha= (|1|)/(|6|)$

$=1/6 :.alpha =tan^-1 (1/6) = 0.165 ; Z_2$ lies on first quadrant,

$:. theta=alpha ~~0.165$ $:. Z_2=6.083(cos 0.165+isin 0.165)$

$Z=(1-2i)/(6+i)$

$Z= (2.236(cos5.176+isin 5.176))/(6.083(cos 0.165+isin 0.165)$

$Z=0.368(cos(5.176-0.165)+isin (5.176-0.165))$ or

$Z=0.368(cos 5.011+isin 5.011) =4/37-13/37i$

In trigonometric form: $0.368(cos 5.011+isin 5.011)$ [Ans]

How do you divide (1-2i) / (6+i) 1−2i6+i (1-2i) / (6+i) in trigonometric form?