How do you divide $\frac{i + 3}{- i + 9}$ $( i+3) / (-i +9 )$ in trigonometric form?

Question

1659 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-15T13:06:45

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

$\frac{3 + i}{9 - i}$ $(3+i)/(9-i)$ $Z = a + i b$ $Z=a+ib$ . 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 :

$Z = | Z | (cos θ + i sin θ); Z_{1} = 3 + i$ $Z =|Z|(costheta+isintheta);Z_1= 3+ i$ .

Modulus: $| Z_{1} | = \sqrt{3^{2} + 1^{2}} \approx 3.16$ $|Z_1|=sqrt(3^2+1^2)~~ 3.16$

Argument: $tan alpha= (|1|)/(|3|):. alpha = tan^-1(1/3)~~0.32$

$Z_1$ lies on first quadrant, so $theta =alpha ~~ 0.32$

$:. Z_1=3.16(cos 0.32+isin 0.32)$

$Z_2= 9 - i$ . Modulus: $|Z_2|=sqrt(9^2+1^2)$

$=sqrt 82~~ 9.06$ Argument: $tan alpha= (|-1|)/(|9|)$

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

quadrant. $:. theta=2pi-alpha ~~6.17$

$:. Z_2=9.06(cos 6.17+isin 6.17) :. (3+i)/(9-i) =$

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

$Z=0.35(cos(0.32-6.17)+isin (0.32-6.17))$ or

$Z=0.35(cos 5.85 +i sin 5.85) =13/41+6/41 i$

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

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