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Question

Let $P(x,y)$ be the point on an Argand diagram representing the complex number $u=x+iy$ and satisfying the equation

\begin{align*} \vert u \vert=k\vert u+a\vert, \end{align*}

where $k$ is a non-negative real number and $a=\alpha+i\beta$ for some real numbers $\alpha$ and $\beta$.

(a) Assume that $k\neq0$ and $k\neq1$. Show that the locus of $P$ is a circle with centre at $\lambda a$ and radius $\vert a\vert\sqrt{\lambda(1+\lambda)}$, where $\lambda$ is the ratio $\dfrac{k^2}{1-k^2}$.

(b) By making the substitution $u=z+b$ and making use of your answer for (a) describe the locus of the complex number $z$ if it satisfies the equation \begin{align*} \vert z+b\vert=k\vert z+c\vert \end{align*} where $b$ and $c$ are fixed complex numbers and $k$ is a non-negative real number equal to neither $0$ nor $1$.

Solution

(a) No problems here.

(b) Is the following correct? Let $u=z+b$. Then the given equation of $\vert z+b\vert=k\vert z+c\vert$ becomes $\vert u\vert=k\vert u+(c-b)\vert$. Hence, by (a), $u$ has a circular locus of radius $\vert c-b\vert\sqrt{\lambda(1+\lambda)}$ and centre $\lambda(c-b)$. But we need the locus of $z=u-b$. This is then a circle with the same radius as that of $u$ but with a centre of $\lambda(c-b)-b.$

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  • $\begingroup$ Your solution seems correct. $\endgroup$ – Emilio Novati Apr 17 '15 at 14:28

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