# Locus of complex numbers $z$ with restricted $(z+1-i)/(z-1-i)$

### Problem

Describe the locus of the following points on the Argand diagram:

$$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1$$ and

$$\mathrm{arg}\left[\frac{(z+1+i)}{(z-1-i)}\right] = \pm \frac{\pi}{2}.$$

### Progress

I've tried putting $z = x+iy$ then rationalising the denominator by multiplying by the conjugate which gave me some numbers, but I'm not sure what to do with them.

I would really love some help on these 2 questions, all responses much appreciated!

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What have you tried? Are you confused with some particular step? It helps for us to know. Regards – Amzoti Apr 1 '13 at 4:12
I've tried putting z = x+iy then rationalising the denominator by multiplying by the conjugate which gave me some numbers, but I'm not sure what to do with them. – Andy Apr 1 '13 at 4:22

to answer the second one $$arg\left({z + 1 + i \over z - 1 -i} \right) = \pm {\pi \over 2}$$ is a circle with diameter $1+i, -1-i.$

here is how you can see this: let $${z + 1 + i \over z - 1 -i} = ki \text{ where k is a real number}$$ you can dove for $z$ and get $$z = {-ki +k - 1 -i\over 1 -ki} = {[k-1 -i(k+1)] \over 1 - ik} = {[k-1 -i(k+1)](1 + ik) \over 1 + k^2} \\= {k-1 + k(k+1) + i[k(k-1)-(k+1)] \over 1 + k^2} = {k^2+ 2k - 1 + i(k^2 - 2k - 1) \over 1 + k^2}$$

if $z = x = iy,$ then $$x = {k^2 - 1 + 2k \over k^2 + 1}, y = {k^2 - 1 - 2k \over k^2 + 1}$$ which gives $$x^2 + y^2 = { 2(k^2 - 1)^2 + 8k^2 \over (k^2 + 1)^2} = { 2(k^2 + 1)^2 \over (k^2 + 1)^2} = 2$$ as claimed at the beginning.

if you look at the geometry of the problem, it says the angle subtended by the point $z$ by the points $\pm(1+i)$ is $90^\circ$ which is the locus of a circle. to derive it algebraically i had to go through all the trouble.

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The first one is saying that the distance between $z$ and $1 + i$ is the same as the distance between $z$ and $1 - i$. The set of points equidistant from two points is the line bisecting the line segment joining the two points. Hence, the locus is the line $y = 0$.

The second one is saying if you look at the point $P$ given by translating $z$ 1 up and 1 to the right and the point $Q$ given by $z$ translated 1 down and 1 to the left, the angle between $P$ and $Q$ (with respect to the origin) is $\pi/2$. Intuitively, this seems to me to be the line $y = x$, though I haven't checked this rigorously.

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Ok thank you for the explanations, so for the first question it's simply the line x=0? Also, for the second question, I thought the angle was always taken between the positive direction of the horizontal - so from which point would we take the angle (does it matter)? – Andy Apr 1 '13 at 4:25
Ooops, I meant $y = 0$, but yes, that's the answer. As for the second one, arg of a quotient is the angle between the numerator and the denominator. (Because multiplying two complex numbers results in the angles that they make with the positive horizontal axis being added) – Jonah Sinick Apr 1 '13 at 4:35

$$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1$$ $$\left|\ (z+1-i)\right| = |(z-1-i)|$$ put z=x+iy $$\left|\ (x+iy+1-i)\right| = |(x+iy-1-i)|$$ $$\left|\ (x+1)+i(y-1)\right| = |(x-1)+i(y-1)|$$ $$\left|\ (x+1)+i(y-1)\right|^2 = |(x-1)+i(y-1)|^2$$ since $$\left|\ (x+iy)\right|^2 = x^2+y^2$$ then $$\left|\ (x+1)+i(y-1)\right|^2 = |(x-1)+i(y-1)|^2$$ $$(x+1)^2+(y-1)^2=(x-1)^2+(y-1)^2$$ $$(x+1)^2=(x-1)^2$$ $$x=0$$ equation of Y-axis.

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