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,
√(a−3)2+b2=√a2+(b+1)2=√(a−1)2+(b−2)2(a−3)2+b2=a2+(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.)
a2−6a+9+b2=a2+b2+2b+1=a2−2a+1+b2−4b+4
Subtract a2+b2 from all equations:
−6a+9=2b+1=−2a+1−4b+4
Solve it just like a normal system of equations:
−6a+9=2b+1⇒3a+b=42b+1=−2a+1−4b+4⇒3a+9b=68b=2b=143a+14=4a=54
Therefore, the solution to this system of equations is 54+14i
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