202212221242 Solution to 1983-CE-AMATH-I-6

The complex number $z$ satisfies the condition

$|z-(3+i)|=|z-(5+5i)|$ .

If $z=x+yi$ , where $x$ and $y$ are real, find and simplify the relation between $x$ and $y$ . Find also the values of $x$ and $y$ for which $|z|$ is a minimum.

Roughwork.

$\begin{aligned} |(x+yi)-(3+i)| & = |(x+yi)-(5+5i)| \\ |(x-3)+(y-1)i| & = |(x-5)+(y-5)i| \\ \sqrt{(x-3)^2+(y-1)^2} & = \sqrt{(x-5)^2+(y-5)^2} \\ (x-3)^2+(y-1)^2 & = (x-5)^2+(y-5)^2 \\ \cdots & = \cdots \\ x+2y-10 & = 0 \\ \end{aligned}$

Substituting $10-2y$ for $x$ ,

$\begin{aligned} |z| (=|x+yi|) & = \sqrt{x^2+y^2} \\ & = \sqrt{(10-2y)^2+y^2} \\ & = \sqrt{5y^2-40y+100} \\ \frac{\mathrm{d}}{\mathrm{d}y}\big( |z|\big) & = \frac{\mathrm{d}}{\mathrm{d}y}\sqrt{5y^2-40y+100} \\ 0 & =: \frac{5y-20}{\sqrt{5y^2-40y+100}} \\ y & = 4 \\ x & = 10-2(4) = 2 \\ \end{aligned}$

Ans. $\begin{pmatrix}x = 2 \\ y = 4\end{pmatrix}$

M	T	W	T	F	S	S
			1	2	3	4
5	6	7	8	9	10	11
12	13	14	15	16	17	18
19	20	21	22	23	24	25
26	27	28	29	30	31