Let the complex number be a+ib, then:
(a+ib)(5+3i)=3-4i.
1/(a+ib)=(a-ib)/(a2+b2), so 5+3i=(a-ib)(3-4i)/(a2+b2).
5+3i=(3a-i(4a+3b)-4b)/(a2+b2).
Equating real and imaginary parts:
(3a-4b)/(a2+b2)=5; (4a+3b)/(a2+b2)=-3.
3a-4b=5(a2+b2); 4a+3b=-3(a2+b2).
Therefore:
(3a-4b)/(4a+3b)=-5/3, 3(3a-4b)=-5(4a+3b),
9a-12b=-20a-15b, 3b=-29a, b=-29a/3,
a2+b2=a2+841a2/9=850a2/9.
4a+3b=4a-29a=-25a=-3(850a2/9)=-850a2/3.
Divide through by -25a (since cannot be zero):
1=34a/3, a=3/34, b=-29a/3=-29/34.
So, the number is (3-29i)/34.
CHECK
(5+3i)(3-29i)/34=(15-136i+87)/34=(102-136i)/34=3-4i. OK