How do you divide $(7-i) / (3-i)$ in trigonometric form?

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Answer 1 · 2018-08-11T11:56:34

In trigonometric form: $2.236(cos 0.18+i sin 0.18)$

$Z=(7-i)/(3-i)$

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

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

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

$Z_1= 7- i$ .Modulus: $|Z_1|=sqrt(7^2+(-1)^2)$

$=sqrt 50 ~~ 7.07$ Argument: $tan alpha= ((|-1|))/(|7|)$

$=1/7 , alpha =tan ^-1 (1/7) ~~ 0.142, Z$ lies on fourth quadrant,

so $theta =2pi-alpha=2pi-0.142 ~~ 6.14$

$:. Z_1=7.07(cos 6.14+i sin 6.14)$ ,

$Z_2= 3- i$ .Modulus: $|Z_2|=sqrt(3^2+(-1)^2)$

$=sqrt 10 ~~ 3.16$ Argument: $tan alpha= ((|-1|))/(|3|)$

$=1/3 , alpha =tan ^-1 (1/3) ~~0.322, Z$ lies on fourth quadrant,

so $theta =2pi-alpha=2pi-0.322 ~~ 5.96$

$:. Z_2=3.16(cos 5.96+i sin 5.96)$ ,

$Z=(7-i)/(3-i)$

$Z= (7.07(cos 6.14+i sin 6.14))/(3.16(cos 5.96+i sin 5.96)$

$Z=2.236(cos(6.14-5.96)+isin (6.14-5.96))$ or

$Z=2.236(cos 0.18+i sin 0.18) =2.2+0.4 i$

In trigonometric form; $2.236(cos 0.18+i sin 0.18)$ [Ans]

How do you divide (7-i) / (3-i) (7-i) / (3-i) in trigonometric form?