Question

How do you divide `( 4i+4) / (6i +5 )` in trigonometric form?

Answer 1

In trigonometric form: `0.725(cos 0.091-isin 0.091)`

Explanation:

`(4+4i)/(5+6i) ;Z=a+ib`. Modulus: `|Z|=sqrt (a^2+b^2)`;

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

`Z =|Z|(costheta+isintheta)` `Z_1= 4+4 i`

Modulus:`|Z_1|=sqrt(4^2+4^2)~~ 5.66`

Argument: `tan alpha= (|4|)/(|4|):. alpha = tan^-1 (1)=0.785`

`Z_1` lies on first quadrant, so `theta =alpha ~~ 0.785`

`:. Z_1=5.66(cos 0.785+isin 0.785)`

`Z_2= 5 + 6i`. Modulus:`|Z|=sqrt(5^2+6^2)`

`=sqrt 61 ~~ 7.81` Argument: `tan alpha= (|6|)/(|5|)`

`=6/5 :.alpha =tan^-1 (1.2) ~~ 0.0876 ; Z_2` lies on first

quadrant.`:. theta=alpha ~~0.876`

`:. Z_2=7.81(cos 0.876+isin 0.876) :. (4+4i)/(5+6i) =`

`Z= (5.66(cos0.785+isin 0.785))/(7.81(cos 0.876+isin 0.876)`

`Z=0.725(cos(0.785-0.876)+isin (0.785-0.876))` or

`Z=0.725(cos (-0.091)+isin (-0.091))` or

`Z=0.725(cos (0.091)-isin (0.091))=(44/61-4/61 i )`

In trigonometric form: `0.725(cos 0.091-isin 0.091)` [Ans]