complex numberslocus
Describe the locus of the following points on the Argand diagram:
$$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1$$ and
$$\mathrm{arg}\left[\frac{(z+1+i)}{(z-1-i)}\right] = \pm \frac{\pi}{2}.$$
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!
Hint: Rewrite as $$\left|z-(-2+2\sqrt{3}i)\right|=2.$$ This "says" that the distance from $z$ to $-2+2\sqrt{3}i$ is $2$. So the locus is the circle with a certain centre, a certain radius.
Draw that circle (crucial). You are interested in lines from the origin to points on your circle. The maximum, minimum angles are at points of tangency. One of them will be obvious from the picture. You can work out the other using once familiar geometry/trigonometry. Note that it is $\text{arg}(z)$ that you are being asked about.
Instead of substituting $z=x+iy$ and making it all messy, here is a simple geometric solution:
As you said, the vectors represented by the points in the numerator and denominator contain the constant angle $\pi\over4$. This means that the locus should be an arc of a circle in which the chord made by joining the 2 given points subtends an angle of $\pi\over4$ on the arc.
Since the chord subtends 90 degrees at the centre, the given points make it very easy to find the centre because it must be directly above (6,3) and directly rightward of (3,6). So it is at (6,6). Then the radius is obviously 3. The cartesian equation for such circle is $$(x-6)^2+(y-6)^2=9$$
But since we want only the part on the right of the chord, we add the additional constraint $$x+y>9$$ (because $x+y=9$ is the equation of the chord.)
These two relations in $x$ and $y$ simultaneously can represent the equivalent of the complex locus given in the Cartesian plane.
Best Answer
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.