Question

How do you divide `(-i+2) / (2i+4)` in trigonometric form?

Answer 1

`(-i+2)/(2i+4)=1/2(costheta+isintheta)`, where `theta=tan^(-1)(-4/3)`

Explanation:

Let us write the two complex numbers in polar coordinates and let them be

`z_1=r_1(cosalpha+isinalpha)` and `z_2=r_2(cosbeta+isinbeta)`

Here, if two complex numbers are `a_1+ib_1` and `a_2+ib_2` `r_1=sqrt(a_1^2+b_1^2)`, `r_2=sqrt(a_2^2+b_2^2)` and `alpha=tan^(-1)(b_1/a_1)`, `beta=tan^(-1)(b_2/a_2)`

Their division leads us to

`{r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)}` or

`{r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)xx(cosbeta-isinbeta)/(cosbeta-isinbeta)}`

`(r_1/r_2){(cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/((cos^2beta+sin^2beta))` or

`(r_1/r_2)*(cos(alpha-beta)+isin(alpha-beta))` or

`z_1/z_2` is given by `(r_1/r_2, (alpha-beta))`

So for division complex number `z_1` by `z_2` , take new angle as `(alpha-beta)` and modulus the ratio `r_1/r_2` of the modulus of two numbers.

Here `-i+2` can be written as `r_1(cosalpha+isinalpha)` where `r_1=sqrt(2^2+(-1)^2)=sqrt5` and `alpha=tan^(-1)(-1/2)`

and `2i+4` can be written as `r_2(cosbeta+isinbeta)` where `r_2=sqrt(4^2+2^2)=sqrt20=2sqrt5` and `beta=tan^(-1)(2/4)=tan^(-1)(1/2)`

and `z_1/z_2=sqrt5/(2sqrt5)(costheta+isintheta)`, where `theta=alpha-beta`

Hence, `tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=((-1/2)-(1/2))/(1+(-1/2)xx(1/2))=(-1)/(3/4)=-4/3`.

Hence, `(-i+2)/(2i+4)=1/2(costheta+isintheta)`, where `theta=tan^(-1)(-4/3)`

Answer 2

`(-i+2)/(2i+4)=1/2(costheta+sintheta)` ,where `theta=tan^-1(-4/3)`

Explanation:

`(-i+2)/(2i+4)`

`=1/2*((2-i)(2-i))/((2+i)(2-i))`

`=1/2*(2^2+i^2-2*2i)/(2^2-i^2)`

`=1/2*(4-1-4i)/(4-(-1))`

`=1/10(3-4i)`

Now `sqrt(3^2+4^2)=5`,So

The given expression

`=1/2(3/5-4/5i)`

Now if we take `3/5=costheta` and `-4/5=sintheta`
i.e.`tantheta=-4/3`
then we can write

`(-i+2)/(2i+4)=1/2(costheta+sintheta)` ,where `theta=tan^-1(-4/3)`