complex......

\hspace{-16}$If $\mathbf{a,b\in \mathbb{R}}$ and $\mathbf{a\neq b}$. Then Locus of all Complex no. $\mathbf{z}$ which\\\\ Satisfy the equation. $\mathbf{\mid z-a \mid^2-\mid z-b \mid^2=1}$

3 Answers

262
Aditya Bhutra ·

putting in z= x+iy and simplifying we get

x= 1+b2-a22(b-a)

thus locus should be Re(z) = 1+b2-a22(b-a)

71
Vivek @ Born this Way ·

Sorry. I misread it.

262
Aditya Bhutra ·

@vivek - which cases ?

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