+506 Let z = a+bi, where a and b are real numbers. After substituting, we get:
\(|(a-3)+bi|=|a+(b+1)i|=|(a-1)+(b-2)i|\)
By the definition of the absolute value of a complex number,
\(\sqrt{(a-3)^2+b^2}=\sqrt{a^2+(b+1)^2}=\sqrt{(a-1)^2+(b-2)^2}\\ (a-3)^2+b^2=a^2+(b+1)^2=(a-1)^2+(b-2)^2\)
(we won't need to worry about extraneous solutions because everything will always be positive since it is the square of real numbers.)
\(a^2-6a+9+b^2=a^2+b^2+2b+1=a^2-2a+1+b^2-4b+4 \)
Subtract \(a^2+b^2\) from all equations:
\(\)\(-6a+9=2b+1=-2a+1-4b+4 \)
Solve it just like a normal system of equations:
\(-6a+9=2b+1\Rightarrow3a+b=4\\ 2b+1=-2a+1-4b+4\Rightarrow 3a+9b=6\\ 8b=2\\ b=\frac{1}{4}\\ 3a+\frac{1}{4}=4\\ a=\frac{5}{4} \)
Therefore, the solution to this system of equations is \(\boxed{\frac{5}{4}+\frac{1}{4}i}\)
