The additive inverses are easy and left as an exercise. But let's try the multiplicative inverse of z = (2,3). In other words, we need to find a complex number (x,y) with the property that (x,y)*(2,3) = (1,0), since we just before found (1,0) to be the multiplicative identity. That's equivalent to:

(x,y)*(2,3) = (2x-3y, 3x+2y) = (1,0)

That results in a linear system of two equations:

(1) 2x-3y = 1
(2) 3x+2y = 0

Multiply the first equation by 2, the second by 3, and add both equations:

4x	- 6y	=	2
9x	+ 6y	=	0
13x		=	2

Thus x = 2/13. Using that in equation (1) gives y = -3/13. You can confirm that indeed (2,3)*(2/13,-3/13)=(1,0)

The remaining questions can be solved similarly and are left to the esteemed reader.