Help!

Hi somebody,

If any of my complex number questions prove to be wrong (or you have reason to question my answers) can you please let me know. :)

Find all complex numbers  z such that

|z - 3| = |z+i| = |z -3i|

$let\;z=a+bi\\ |(a-3)+bi|=|a+(b+1)i|=|a+(b-3)i|\\ (a-3)^2+b^2=a^2+(b+1)^2=a^2+(b-3)^2\\ a^2-6a+9+b^2=a^2+b^2+2b+1=a^2+b^2-6b+9\\ -6a+9=2b+1=-6b+9\\~\\ 2b+1=-6b+9\\\\ 8b=8\\ b=1\\~\\ -6a+9=3\\ -6a=-6\\ a=1$

So I get     

$z=1+i$

Yah Melody thats what i got

Yes, I'm pretty sure that is right. Thank you sooo much tho!!! i finally figured out that u have to substitute z = a+bi. THanks again!!!