Cierra Castillo

Answered

2022-07-12

If argument of $\frac{z-{z}_{1}}{z-{z}_{2}}$ is $\frac{\pi}{4}$, find the locus of $z$.

${z}_{1}=2+3i$

${z}_{2}=6+9i$

Approach: I tried to solve the equation using diagram, basically plotting the points on the Argand plane. What I got is a circle with center $7+4i$ and a radius of $\sqrt{26}$ units. The two complex numbers given lie on this circle, and form a chord. Any point lying on the major arc of this chord satisfies the condition.

How exactly would I represent this as a locus of the point? And is there any other method that I can use that does not involve a diagram?

${z}_{1}=2+3i$

${z}_{2}=6+9i$

Approach: I tried to solve the equation using diagram, basically plotting the points on the Argand plane. What I got is a circle with center $7+4i$ and a radius of $\sqrt{26}$ units. The two complex numbers given lie on this circle, and form a chord. Any point lying on the major arc of this chord satisfies the condition.

How exactly would I represent this as a locus of the point? And is there any other method that I can use that does not involve a diagram?

Answer & Explanation

Jenna Farmer

Expert

2022-07-13Added 17 answers

the angle subtended by the chord ${z}_{1}{z}_{2}$ at the center is $2\pi /4=\pi /2$ so the radius is $\frac{|{z}_{1}-{z}_{2}|}{\sqrt{2}}=\sqrt{26}$ the center of the chord is $4+3i$ you add or subtract $\frac{-6+4i}{2}$ so that you will get two centers. the two centres, ${z}_{1}$ and ${z}_{2}$ form a square of side $\sqrt{26}.$

Savanah Boone

Expert

2022-07-14Added 5 answers

Put $\phantom{\rule{thickmathspace}{0ex}}z=x+iy\phantom{\rule{thickmathspace}{0ex}},\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x,y\in \mathbb{R}\phantom{\rule{thickmathspace}{0ex}}$, so

$\frac{z-2-3i}{z-6-9i}=\frac{(x-2)+(y-3)i}{(x-6)+(y-9)i}\cdot \frac{(x-6)-(y-9)i}{(x-6)-(y-9)i}=$

$=\frac{(x-2)(x-6)+(y-3)(y-9)}{(x-6{)}^{2}+(y-9{)}^{2}}+\frac{(x-6)(y-3)-(x-2)(y-9)}{(x-6{)}^{2}+(y-9{)}^{2}}i$

By the given data, it must be that the real and imaginary parts are identical, and thus

$(x-2)(x-6)+(y-3)(y-9)=(x-6)(y-3)-(x-2)(y-9)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-14x+{y}^{2}-8y-26=0$

Complete squares, make some algebraic hokus pokus and get a circle.

$\frac{z-2-3i}{z-6-9i}=\frac{(x-2)+(y-3)i}{(x-6)+(y-9)i}\cdot \frac{(x-6)-(y-9)i}{(x-6)-(y-9)i}=$

$=\frac{(x-2)(x-6)+(y-3)(y-9)}{(x-6{)}^{2}+(y-9{)}^{2}}+\frac{(x-6)(y-3)-(x-2)(y-9)}{(x-6{)}^{2}+(y-9{)}^{2}}i$

By the given data, it must be that the real and imaginary parts are identical, and thus

$(x-2)(x-6)+(y-3)(y-9)=(x-6)(y-3)-(x-2)(y-9)\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}$

$\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}{x}^{2}-14x+{y}^{2}-8y-26=0$

Complete squares, make some algebraic hokus pokus and get a circle.

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