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How would I sketch the region: $|z + 1| = 4|z - 1|$?

I decided to manipulate the expression and square both sides of the expression to get the follow. I think I am on to something, compared to what I had previously at least.

\begin{eqnarray} |z + 1| & = & 4|z - 1|\\ \sqrt{(x+1)^2+y^2} & = & 4\sqrt{(x-1)^2 + y^2}\\ \Rightarrow (x+1)^2 + y^2 & = & 16[(x-1)^2 + y^2]\\ x^2 + 2x + 1 + y^2& = &16(x^2 -2x + 1) + 16y^2 \\ \Rightarrow x^2 + 2x + 1 -16x^2 + 32x - 16 & = & 15y^2\\ -15x^2 + 34x - 15& = & 15y^2\\ -x^2 + \frac{34}{15}x - 1 & = & y^2\\ -\left(x^2 - \frac{34}{15}x\right) - 1 & = & y^2\\ -\left(x^2 -\frac{34}{15}x + \frac{289}{225}\right) + \frac{289}{225}- 1& = & y^2 \\ \Rightarrow \frac{64}{225} & = & \left(x - \frac{17}{15}\right)^2 + y^2\\ \Rightarrow \frac{8}{15} & = & \sqrt{\left(x- \frac{17}{15}\right)^2 + y^2}\\ & = & \left|z - \frac{17}{15}\right| \end{eqnarray}

From what I get, the region that I would sketch would be a circle centered at the point $z = \frac{17}{15}$ with a radius of $\frac{8}{15}$.

Is this a reasonable conclusion?

Thank you for your time and thanks in advanced for any feedback.

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Please square both sides. –  Erdos Yi May 10 at 3:03
    
Well they are norms/magnitudes/modulus expression. If I am not mistaken, when you square a modulus expression it is not the same as squaring something like (x + y). –  Kevin_H May 10 at 3:09
    
The reason for squaring the absolute value is that $|a+bi|^2=a^2+b^2$, since $|z|^2=z\overline z$. –  Lubin May 10 at 3:16

1 Answer 1

Call $z=x+iy$; then you have \begin{align*} |z+1|=4|z-1|\Longleftrightarrow& |z+1|^2=16|z-1|^2\\ \Longleftrightarrow&(x+1)^2+y^2=16(x-1)^2+16y^2\\ \Longleftrightarrow&x^2+2x+1+y^2=16x^2-32x+16+16y^2\\ \Longleftrightarrow&15x^2-34x+15y^2+15=0\\ \Longleftrightarrow&x^2-\frac{34}{15}x+y^2+1=0 \end{align*}

which is an ellipse.

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