how do i solve such a thing

http://img353.imageshack.us/img353/672/85253506or3.gif
Any number of ways. First, note that you really have two equations there. There may not exist z satisfying both of them.

For one of them, say |z+ i|= |z- 1|, let z= x+ iy and expand it: z+ i= x+ (y+1)i and |z+ i|= [itex]\sqrt{x^2+ (y+1)^2}[/itex] while z-1= (x- 1)+ iy so |z-1|= [itex]\sqrt{(x-1)^2+ y^}[/itex] |z+ i|= |z- 1|, then, is the same as [itex]\sqrt{x^2+ (y+1)^2}= \sqrt{(x-1)^2+ y^2}[/itex] so [itex]x^2+ (y+1)^2= (x-1)^2+ y^2[/itex]. If you multiply those out the squares cancel and you should see that is a straight line in the xy-plane. You could do the same thing with [itex]|z+1|= |1/\sqrt{2}+ i/\sqrt{2}- z| and again see the it is a straight line in the xy-plane. The solution of the entire system is the point where those two lines intersect.

Of you could do it geometrically: |a- b| can be interpreted as the distance between points a and b in the complex plane. In particular, |z-1|= |z+ i| is the set of points that are equidistant from -1 and i or, in terms of (x,y), (-1, 0) and (0, 1). That is, of course, the perpendicular bisector on that line segement. Similarly, [itex]|z-1|= |1/\sqrt{2}+ i/\sqrt{2}- z| is the set of points equidistant from 1 and [itex]1/\sqrt{2}+ i/\sqrt{2}[/itex]: the perpendicular bisector of the line segment from (1, 0) to [itex](1/\sqrt{2}, 1/\sqrt{2})[/itex]. The point satisfying the original system is the point where those two lines intersect.