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Here's a question I'm working on:

Suppose that $k>0$ and $z_1, z_2 \in \mathbb{C}$
1). Show that $|z - z_1| = k \cdot |z - z_2|$ determines a circle if $k \neq 1$
2). Describe the locus of the set of points if $k=1$

I'm a bit confused on the first question, and not sure where to start. Any help would be appreciated.

Thanks.

HoopsMcCann
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  • Hint for 1) square both sides, express them in terms on $(x,y)$ and compute LHS - RHS. What is the general form of the result? does that match the one for a circle? – achille hui May 25 '17 at 15:04

1 Answers1

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\begin{align} (z-z_1)(\bar z-\bar z_1)&=k^2(z-z_2)(\bar z-\bar z_2)\\ (k^2-1)z\bar z-(k^2z_2-z_1)\bar z-(k^2\bar z_2-\bar z_1)z&=z_1\bar z_1-k^2z_2\bar z_2 \end{align}

If $k\ne 1$, let $\displaystyle \beta=\frac{k^2z_2-z_1}{k^2-1}$. The above equation can be written as

\begin{align} z\bar z-\beta\bar z-\bar \beta z+\beta\bar \beta&=\frac{z_1\bar z_1-k^2z_2\bar z_2}{k^2-1}+\beta\bar\beta\\ (z-\beta)(\bar z-\bar \beta)&=\frac{z_1\bar z_1-k^2z_2\bar z_2}{k^2-1}+\beta\bar\beta\\ |z-\beta|^2&=\frac{z_1\bar z_1-k^2z_2\bar z_2}{k^2-1}+\beta\bar\beta\\ \end{align}

Note that

\begin{align} \frac{z_1\bar z_1-k^2z_2\bar z_2}{k^2-1}+\beta\bar\beta&=\frac{z_1\bar z_1-k^2z_2\bar z_2}{k^2-1}+\frac{(k^2z_2-z_1)(k^2\bar z_2-\bar z_1)}{(k^2-1)^2}\\ &=\frac{(k^2-1)z_1\bar z_1+(k^2-k^4)z_2\bar z_2}{(k^2-1)^2}\\ &\qquad +\frac{k^4z_2\bar z_2-k^2z_1\bar z_2-k^2\bar z_1 z_2+z_1\bar z_1}{(k^2-1)^2}\\ &=\frac{k^2(z_1-z_2)(\bar z_1-\bar z_2)}{(k^2-1)^2} \end{align}

The eequation can be written as

\begin{align} |z-\beta|^2&=\frac{k^2|z_1-z_2|^2}{(k^2-1)^2}\\ |z-\beta|&=\frac{k|z_1-z_2|}{|k^2-1|} \end{align}

and represents a circle.

CY Aries
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