Author Topic: Q1: TUT 0301  (Read 829 times)

Victor Ivrii

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Q1: TUT 0301
« on: September 28, 2018, 04:15:39 PM »
$\renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}$
Describe the locus of points $z$ satisfying the given equation.
\begin{equation*}
|z-1|^2=|z+1|^2+6.
\end{equation*}

Meng Wu

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Re: Q1: TUT 0301
« Reply #1 on: September 28, 2018, 04:34:22 PM »
Let $$z=x+iy,\text{where} \space x,y \in \mathbb{R}.$$Given equation becomes:
$$|x+iy-1|^2=|x+iy+1|^2+6$$
Since,$$|x+iy-1|=\sqrt{(x-1)^2+y^2} \\ |x+iy+1|=\sqrt{(x+1)^2+y^2}$$
Hence, we have$$\require{cancel}\begin{align}\bigg(\sqrt{(x-1)^2+y^2}\bigg)^2&=\bigg(\sqrt{(x+1)^2+y^2}\bigg)^2+6 \\(x-1)^2+y^2&=(x+1)^2+y^2+6 \\ \cancel{x^2}-2x+\cancel{y^2}&=\cancel{x^2}+2x+\cancel{y^2}+6\\-4x&=6\\x&=-\frac{3}{2}\end{align}$$
$\\$
$\\$
Therefore, the locus of points $z$ are the straight line(vertical line) $x=-\frac{3}{2}.$
« Last Edit: September 28, 2018, 04:36:07 PM by Meng Wu »

Meng Wu

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Re: Q1: TUT 0301
« Reply #2 on: September 28, 2018, 05:51:56 PM »
Just notice, I posted the answer before the 18:00 you mentioned. Is this gonna be an issue? Just wondering.