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
