(1+3i)(2-5i)=2+15+i=17+i.
Assuming (2-6^1/2) is supposed to be (2-6^1/2i),
(2-i√6)(-3+2i√5)=-6+i(4√5+3√6)+2√30.
The reciprocal of this is:
[(2√30-6)-i(4√5+3√6)]/[(2√30-6)²+(4√5+3√6)²]=
[(2√30-6)-i(4√5+3√6)]/(120+36-24√30+80+54+24√30)=
[(2√30-6)-i(4√5+3√6)]/290.
So the result of the division is (17+i)[(2√30-6)-i(4√5+3√6)]/290.
The real part of this is (34√30-102+4√5+3√6)/290;
The imaginary part is -68√5-51√6+2√30-6)/290.
The modulus is the square root of the sum of the squares of the imaginary and real components. It would be impractical to give the results in radicals, so the numerical values will be calculated.
The real part is 0.3466 approx and the imaginary part is -0.9380 approx.
The sum of the squares is 1 and the modulus is therefore 1.
A BETTER APPROACH
We can write the quotient as z₁/z₂ where z₁=r₁(cosθ+isinθ) and z₂=r₂(cosɸ+isinɸ), where r₁ and r₂ are the moduli of z₁ and z₂. So z₁=17+i and z₂=(2√30-6)+i(4√5+3√6). We don’t need to work out the angles.
From these, r₁=√(17²+1)=√290 and r₂=√(120+36-24√30+80+54+24√30)=√290.
Therefore r₁=r₂. Let r=r₁=r₂, then z₁=r(cosθ+isinθ) and z₂=r(cosɸ+isinɸ).
The reciprocal of z₂ is z₃=(1/r)(cosɸ-isinɸ). Let z=z₁z₃=cos(θ-ɸ)+isin(θ-ɸ).
The modulus of z is 1, so the modulus of the quotient z₁/z₂ is 1.