# Show that the real part of complex number is $-1$

The complex number z satisfies the equation $$\vert z \vert=\vert z+2\vert$$. Show that the real part is $$-1$$.

I know that $$\vert z \vert = \sqrt{x^2+y^2}$$ so I took

\begin {align}\vert z \vert&=\vert z+2\vert \\\sqrt{x^2+y^2} &=\sqrt {x^2+y^2+2^2} \\ x^2+y^2&=x^2+y^2+2^2 \end {align}

So after cancelling $$x^2$$ and $$y^2$$ from both sides of the equation, I am left wih $$0=2^2$$, which makes no sense.

How should I solve this question?

The second part of the question is as follows (which I also need help solving):

The complex number $$z$$ also satisfies $$\vert z \vert -3=0$$. Represent the two possible values of $$z$$ in an Argand diagram. Calculate also the two possible values of arg $$z$$.

It is not true that $$|z+2|^{2}=x^{2}+y^{2}+2$$. What is true is $$|z+2|^{2}=|(x+2)+iy)|^{2}=(x+2)^{2}+y^{2}=x^{2}+y^{2}+2^{2}+4x$$. So we get $$4x+4=0$$ or $$x=-1$$.

$$|z + 2| = \sqrt{(x+2)^2 + y^2} = \sqrt{x^2 + y^2}$$ $$x^2 + 4x + 4 = x^2$$

• I believe you meant $4x$ instead of $2x$ Mar 20, 2020 at 6:02
• Oops sorry, fixed! Mar 20, 2020 at 6:03

I believe there's a mistake when you start expanding the modulus expressions. In particular, $$|z+2| = |x+iy + 2| = \sqrt{(x+2)^2 + y^2}$$. Try using this and see if it fixes your results.

So, $$z$$ are the intersections of two circle with equal radius centered at $$(0,0)$$ and $$(-2,0)$$. Quite straightforward that $$z$$ lies on the line $$x=-1$$.

Part two basically tell You the radius is $$3$$, intersections are $$-1\pm i2\sqrt{2}$$

You’ve miscalculated the length of $$z+2$$.

Equal the squares of the lengths to obtain $$z\bar z=(z+2)(\overline{z+2})$$ hence $$z+\bar z=2,$$ which proves the statement.