To convert non-zero complex number #a+ib# into trigonometric form #r(cos alpha + i sin alpha)#
we have to multiply and divide it by #sqrt(a^2+b^2)# getting
#sqrt(a^2+b^2)(a/sqrt(a^2+b^2)+i b/sqrt(a^2+b^2))#
Now there is always one and only one angle #alpha# such that
#cos alpha = a/sqrt(a^2+b^2)# and
#sin alpha = b/sqrt(a^2+b^2)#
So, in trigonometric form our number would look like
#sqrt(a^2+b^2)(cos alpha + i sin alpha)#
where angle #alpha# is defined by its #cos# and #sin# as explained above.
Furthermore, trigonometric form #cos alpha +i sin alpha# is, using the Euler's formula, equivalent to #e^(i alpha)#, which will make it easy to multiply and divide complex numbers.
In our problem we have, using this logic,
#4i+1 = sqrt(17)(cos phi + i sin phi) = sqrt(17)e^(i phi)#
where #phi = arccos(1/sqrt(17))~~75.96^o#
#6i+5 = sqrt(61)(cos psi + i sin psi) = sqrt(61)e^(i psi)#
where #psi = arccos(5/sqrt(61))~~50.19^o#
Therefore,
#(4i+1)/(6i+5) = sqrt(17)/sqrt(61)e^(i phi)/e^(i psi) #
#= sqrt(17/61) e^(i(phi-psi))#
#=sqrt(17/61)(cos(phi-psi)+i sin(phi-psi))#
#~~sqrt(17/61)(cos(25.77^o)+i sin(25.77^o))#
#~~0.5279(0.9005+i*0.4347)#
#~~0.4754+i*0.2295#
