# How do you divide ( i+3) / (-i +9 ) in trigonometric form?

Feb 15, 2018

In trigonometric form: $0.35 \left(\cos 5.85 + i \sin 5.85\right)$

#### Explanation:

$\frac{3 + i}{9 - i}$ $Z = a + i b$. Modulus: $| Z | = \sqrt{{a}^{2} + {b}^{2}}$;

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

Z =|Z|(costheta+isintheta);Z_1= 3+ i .

Modulus:$| {Z}_{1} | = \sqrt{{3}^{2} + {1}^{2}} \approx 3.16$

Argument: $\tan \alpha = \frac{| 1 |}{| 3 |} \therefore \alpha = {\tan}^{-} 1 \left(\frac{1}{3}\right) \approx 0.32$

${Z}_{1}$ lies on first quadrant, so $\theta = \alpha \approx 0.32$

$\therefore {Z}_{1} = 3.16 \left(\cos 0.32 + i \sin 0.32\right)$

${Z}_{2} = 9 - i$. Modulus:$| {Z}_{2} | = \sqrt{{9}^{2} + {1}^{2}}$

$= \sqrt{82} \approx 9.06$ Argument: $\tan \alpha = \frac{| - 1 |}{| 9 |}$

=1/9 :.alpha =tan^-1 (1/9) = 0.11 ; Z_2 lies on fourth

quadrant.$\therefore \theta = 2 \pi - \alpha \approx 6.17$

$\therefore {Z}_{2} = 9.06 \left(\cos 6.17 + i \sin 6.17\right) \therefore \frac{3 + i}{9 - i} =$

 Z= (3.16(cos0.32+isin 0.32))/(9.06(cos 6.17+isin6.17)

$Z = 0.35 \left(\cos \left(0.32 - 6.17\right) + i \sin \left(0.32 - 6.17\right)\right)$ or

$Z = 0.35 \left(\cos 5.85 + i \sin 5.85\right) = \frac{13}{41} + \frac{6}{41} i$

In trigonometric form; $0.35 \left(\cos 5.85 + i \sin 5.85\right)$