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

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)#

#(-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)#