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

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Answer 1 · 2018-01-30T12:42:16

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

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

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

Argument: $theta=tan^-1(b/a)$ Trigonometrical form :

$Z =|Z|(costheta+isintheta)$

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

$=sqrt 5 ~~ 2.236$ Argument: $tan alpha= (|-2|)/(|1|)$

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

$theta =2pi-alpha=2pi-1.107 ~~ 5.176$

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

$Z_2= 6 + i$ .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) (1-2i) / (6+i) in trigonometric form?