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

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Answer 1 · 2018-08-11T11:56:34

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

$Z = \frac{7 - i}{3 - i}$ $Z=(7-i)/(3-i)$

$Z = a + i b$ $Z=a+ib$ . 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 :

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

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

$= \sqrt{50} \approx 7.07$ $=sqrt 50 ~~ 7.07$ Argument: $tan α = \frac{(| - 1 |)}{| 7 |}$ $tan alpha= ((|-1|))/(|7|)$

$= \frac{1}{7}, α = {tan}^{- 1} (\frac{1}{7}) \approx 0.142, Z$ $=1/7 , alpha =tan ^-1 (1/7) ~~ 0.142, Z$ lies on fourth quadrant,

so $θ = 2 π - α = 2 π - 0.142 \approx 6.14$ $theta =2pi-alpha=2pi-0.142 ~~ 6.14$

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

$Z_2= 3- i$ .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−i3−i (7-i) / (3-i) in trigonometric form?