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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 $z$.

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5 Answers 5

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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 $x=-1$.

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

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    $\begingroup$ I believe you meant $4x$ instead of $2x$ $\endgroup$
    – gc3941d
    Mar 20, 2020 at 6:02
  • $\begingroup$ Oops sorry, fixed! $\endgroup$ Mar 20, 2020 at 6:03
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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.

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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 $-1\pm i2\sqrt{2}$

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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.

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