Question

How do you divide `( i-3) / (5i +2)` in trigonometric form?

Answer 1

`(i-3)/(5i+2)~~0.587*(cos(1.630)+i * sin(-1.630))`

Explanation:

`color(red)("~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~")`
`color(white)("XX ")color(blue)(a+bi harr r * [cos(theta)+i * sin(theta)])`
`color(white)("XXX")color(blue)("where " r=sqrt(a^2+b^2))`
`color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})`
`color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")`

`color(red)("~~~ Trigonometric Division ~~~")`
#color(white)("XX ")color(blue)((cos(theta)+i * sin(theta))/(cos(phi)+i * sin(phi)))#

`color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta) * cos(phi)-cos(theta) * sin(phi))]`

`color(red)("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")`

If `A=i-3=-3+1i`
then
`color(white)("XXX")r_A=sqrt((-3)^2+1^2)=sqrt(10)`
and
`color(white)("XXX")theta_A=arctan(-1/3)+pi~~2.819842099`

If `B=5i+2=2+5i`
then
`color(white)("XXX")r_B=sqrt(2^2+5^2)=sqrt(29)`
and
`color(white)("XXX")theta_B=arctan(5/2)~~1.19028995`

`A/B=(i-3)/(5+2i)`

`color(white)("XXX")=sqrt(10)/sqrt(29) * ([cos(theta_A)+isin(theta_A))/(cos(theta_B)+isin(theta_B)))`

`color(white)("XXX")=sqrt(10)/sqrt(29) *([color(green)(cos(theta_A)*cos(theta_B)+sin(theta_A) * sin(theta_B))+i[color(magenta)(sin(theta_A) * cos(theta_B) - cos(theta_A) * sin(theta_B))])`

Plugging in the values for `theta_A` and `theta_B` and using a calculator/spreadsheet:
`color(white)("XXX")~~-0.034482759+0.586206897i`

If we call this value `C`
then
`color(white)("XXX")r_C=sqrt((-0.03448...)^2+(0.5862...)^2)~~0.58722022`
and (since `C` is in Quadrant 2)
`color(white)("XXX")theta_C=arctan((0.5862...)/(-0.03448...))+pi~~1.62955215`

`A/B=C=r_C(cos(theta_C)+i * sin(theta_C))`

Answer 2

`(i-3)/(5i+2)~~0.587(cos(1.630)+i * sin(1.630))`

Explanation:

This is an alternate (and I think simpler solution) than the first one given (probably below this one) but it does not do the division in trigonometric form; instead, it does the division and then converts the result into trigonometric form.

`(i-3)/(5i+2) = ((i-3) * (5i-2))/((5i+2) * (5i-2))`

#color(white)("XXXXXXXXX"){:(underline(xx),underline("|"),underline(i),underline(-3)),(5i,"|",-5,-15i),(underline(-2),underline("|"),underline(-2i),underline(+6)),(,,1,-17i):} color(white)("XXXX"){:(underline(xx),underline("|"),underline(5i),underline(+2)),(5i,"|",-25,+10i),(underline(-2),underline("|"),underline(-10i),underline(-4)),(,,-29,):}#

`(i-3)/(5i+2)=(1-17i)/(-29)=-1/29+i * 17/29`

If we call this value `C`
then
`color(white)("XXX")r_C=sqrt((1/29)^2+(17/29)^2)~~0.58722022`
and (since `C` is in Quadrant II)
`color(white)("XXX")theta_C=arctan(17/(-1))+pi~~1.62955215`