To convert non-zero complex number a+ib into trigonometric form r(cosα+isinα)
we have to multiply and divide it by √a2+b2 getting
√a2+b2(a√a2+b2+ib√a2+b2)
Now there is always one and only one angle α such that
cosα=a√a2+b2 and
sinα=b√a2+b2
So, in trigonometric form our number would look like
√a2+b2(cosα+isinα)
where angle α is defined by its cos and sin as explained above.
Furthermore, trigonometric form cosα+isinα is, using the Euler's formula, equivalent to eiα, which will make it easy to multiply and divide complex numbers.
In our problem we have, using this logic,
4i+1=√17(cosϕ+isinϕ)=√17eiϕ
where ϕ=arccos(1√17)≈75.96o
6i+5=√61(cosψ+isinψ)=√61eiψ
where ψ=arccos(5√61)≈50.19o
Therefore,
4i+16i+5=√17√61eiϕeiψ
=√1761ei(ϕ−ψ)
=√1761(cos(ϕ−ψ)+isin(ϕ−ψ))
≈√1761(cos(25.77o)+isin(25.77o))
≈0.5279(0.9005+i⋅0.4347)
≈0.4754+i⋅0.2295
