If these two equations are simultaneously true, we can square equation (b): (2-j3)^2=a+jb=(2+j)(3-j2). It's not clear whether j2 is j^2 or 2j or jsub2, so we'll just use its identity as it stands for the moment. It may even be that j=i=imaginary square root of -1 and can be used instead of i to avoid confusion with i for electric current.
4-4j3+(j3)^2=6-2j2+3j-j(j2).
(A) Assume j2=j^2, j3=j^3: j^6-3j^3+2j^2-3j-2=0. This doesn't look like the right interpretation.
(B) Now assume j is another way writing i. So j2=j^2=-1, j3=j^3=-j, j^4=1, j^6=-1.
That gives us: 4+4j-1=6+2+3j+j; 3+4j=8+4j; 3=8, which is not true so j is not i, if (a) and (b) are simultaneously true.
(C) Assume j=i but (a) and (b) are independent equations.
(a) 8+4j=a+jb so a=8 and b=4
(b) 3+4j=a+jb so a=3 and b=4; 2-j3=2-j^3=2+j=(a+jb)^0.5=(3+4j)^0.5
I think (C) is the most likely answer.