Find all complex numbers $z$ such that $Re((z−1)^2)= 0$ and $|z+1|=2$

Question

How do you find all complex numbers $z$ such that $Re((z−1)^2)= 0$ and $|z+1|=2$

I don't know where should I start since I only know the real part.

Answer 1

With $$z=x+iy$$ we get $$\Re((x-1)^2-y^2+2y(x-1)i)=(x-1)^2-y^2=0$$ and $$\sqrt{(x+1)^2+y^2}=2$$

Answer 2

First, $|z+1|=2$ is the equation of a circle centered at $\Omega(-1)$. Then, if $z=x+iy$, $(x,y)\in{\mathbb R}^2$, $$(z-1)^2=(x-1)^2-y^2+2i(x-1)y$$ so ${\rm Re}((z-1)^2)=0$ is equivalent to $$(x-1)^2-y^2=0\iff (x-1-y)(x-1+y)=0\iff y=x-1\text{ or }y=-x+1$$ Draw the circle, the two lines, figure out the position of the intersections. Or substitute $y$ in the circle equation to find the corresponding $x$...

Answer 3

Let $z = x + iy$. Then $z^2 = x^2 - y^2 + 2xyi$

$$(z-1)^2 = z^2 - 2z + 1 \\ = (x^2 - y^2 + 2xyi) - 2(x + iy) + 1 = (x^2 - 2x - y^2 + 1) + i(2xy - 2y)$$

So $Re((z-1)^2) = x^2 - 2x - y^2 + 1 = 0$, which is a pair of intersecting lines.

$$|z+1|^2 = |(x + 1) + iy|^2 \\= (x+1)^2 + y^2 = 4$$

which is a circle of radius 2 centered at $(-1, 0)$

The set you desire will be the intersection of these objects.