The equation $|z-\omega|^2+|z-\omega^2|^2=\lambda$ will represent a circle if

Question

If $\omega$ is a complex cube root of unity and $\lambda$ belongs to real numbers, then the equation $$|z-\omega|^2+|z-\omega^2|^2=\lambda$$ will represent the circle for what range of $\lambda$?

My Attempt:

$\Rightarrow 2|z|^2+|\omega|^2+|\omega^2|^2-(z\omega^2+\bar z\omega+z\omega+\bar z\omega^2)=\lambda$

$\Rightarrow 2|z|^2+2-(\omega^2)(z+\bar z)-(\omega)(z+\bar z)=\lambda$

$\Rightarrow 2(|z|^2+1)+(z+\bar z)=\lambda$

$\Rightarrow 2[|z|^2+Re(z)+1]=\lambda$

$\Rightarrow x^2+y^2+x+1-\lambda/2=0$

For a general conic to be a circle, the determinant must be non zero, coefficient of $xy$ must $=0$, and coefficient of $x^2$ and $y^2$ must be $=1$. Conditions $2$ and $3$ are satisfied, so I checked for condition $1$,and got the answer as $\lambda$ belongs to the real numbers except $-{3/2}$. However the answer is $[3/2,\infty]$. I have already checked with another method and the given answer seems to be correct (involves locus of circle in complex form for diametric endpoints)

Where have I gone wrong?

Answer 1

We have that

$$x^2+y^2+x+1-\frac \lambda 2=0\iff\left(x+\frac12\right)^2+y^2=\frac \lambda 2-\frac 34$$

with the condition $\frac \lambda 2-\frac 34\ge 0 \iff \lambda \ge \frac 32$.

Answer 2

Just another similar way, from your equation $2(|z|^2+1)+(z+\bar z)=\lambda $, it is also possible to write directly that $$\tfrac12\lambda = |z|^2 + \tfrac12z + \tfrac12\bar z+1 = (z +\tfrac12 )(\bar z +\tfrac12 )+\tfrac34$$ $$\iff \left|z +\tfrac12\right|^2 = \frac{2\lambda-3 }4$$

Now that's the equation of a (non-degenerate) circle entered at $-\frac12$, iff RHS is positive, i.e. $\lambda > \frac32$.

Answer 3

The locus of points $P$ on a plane such that $PA^2+PB^2= k$ for fixed $A,B$ is a circle centered at $D$ the mid-point of $AB$

Proof: We know that $PA^2+PB^2 = 2PD^2+\dfrac{AB^2}{2}$ and hence $PD^2 = \dfrac{k-\dfrac{AB^2}{2}}{2}$ i.e. a constant.

Hence for the circle to be non-degenerate we need $k>\dfrac{AB^2}{2}$ i.e. $\lambda > \dfrac{|\omega-\omega^2|^2}{2}=\dfrac{3}{2}$