Simple way of finding the locus of $z$?

Question

I am tasked to sketch the locus of complex numbers $z$ that satisfies $$(z-2+i)^2 + (z^*-2-i)^2 + 2zz^* + 6 = 0$$ I was wondering if there’s a special trick to this question, or if this is only solvable by plugging in $z=x+iy$ ?

I noticed that the second term is the conjugate of the first, and wanted to substitute $w = z-2+i$, however this gives me a complicated expression for $zz^*$ which makes it more confusing. Is there a trick I’m missing out? Thanks!

Edit: I have already solved it through direct substitution (parabola intersecting real axis at $x=3$ and $x=-1$) but it took a good 5 minutes or so, expanding everything correctly, which seems rather long (I was attempting it under exam conditions). I was just wondering if it is possible to exploit the $w$ substitution I made in particular.

Answer 1

Put $z=x+yi$

The expression turns into $$[(x-2)+i(y+1)]^2+[(x-2)-i(y+1)]^2+2(x^2+y^2)+6$$ $$=2(x^2-4x+4)-2(y^2+2y+1)+2(x^2+y^2)+6$$

Which simply turns out to be $$4x^2-8x+12-4y=0$$ $$\Rightarrow y=x^2-2x+3$$

Which is clearly a parabola

Answer 2

You know the answer is quadratic in $x$ and $y$. Can you quickly figure out the coefficients of $x^2$ and $y^2$? Once you do, you'll realize that the locus has a pretty simple form.

That being said: in general, there's no quick and easy method for doing problems like this, and just multiplying out and then completing the square may be your best bet as a general skill.

Answer 3

HINT

Note that

$$(z-2+i)^2 + (z^*-2-i)^2 + 2zz^* + 6 = 0\iff (z+\bar z)^2-4(z+\bar z)+2i(z-\bar z)+12=0$$

then use

• $z+\bar z=2\,\Re(z)$
• $z-\bar z=2i\,\Im(z)$