If $z$ lies on the circle $|z-1|=1$ then$\frac {z-2} z$ is a purely imaginary number

Question

If $z$ lies on the circle $|z-1|=1$ then $\frac {z-2} z$ is a purely imaginary number. This is what by book states. Did'nt understand why. Can someone help?

Actually I was thinking of a more geometrical approach to the problem, as pointed out by @did and the answer below. I also remembered the method that if $\bar z = -z$ then $z$ is purely imaginary. But can it also be done by visual geometry? Ok, I drew a circle centred at $1$ with radius $1$. The statement given implies $\frac {z-2} z$ equals any point on the semicircle. After that how do we prove its imaginary from there?

$$z=1+e^{it}\implies\frac{z-2}z=\frac{e^{it}-1}{e^{it}+1}=\cdots$$ — Did, Sep 16 '15 at 09:37
Consider an $x+yj$ that satisfies $|z-1|=1$ - then work out $\frac{z-2}{z}$ - that's it — Alec Teal, Sep 16 '15 at 09:37
@Did while right, it's unlikely the OP understands the $re^{j\theta}$ form. Should anyone answer please bear this in mind. — Alec Teal, Sep 16 '15 at 09:37
@AlecTeal Difficult to say since the OP posts only PSQs (but some of them were already about complex numbers, so we'll see...). // Later on: "dont worry..i do" OK, we saw... — Did, Sep 16 '15 at 09:40
By putting anything personal in the pot. As you are probably well aware... — Did, Sep 16 '15 at 09:51
@SanchayanDutta I've tried to answer it on geometrical lines. — Shailesh, Sep 16 '15 at 10:33
@SanchayanDutta Thanks. It always helps to think of complex numbers and the various operations geometrically to the extent possible. — Shailesh, Sep 16 '15 at 10:37
(Up to a change of variable, but I'd say it's close enough). — Arnaud D., Aug 29 '18 at 13:05
Does this answer your question? complex numbers : If z lies on the circle $|z-1|=1,$ then the value of $\frac{z-2}{z}$ is ........... — UmbQbify, Aug 02 '20 at 03:22

Sam Cappleman-Lynes · Answer 1 · 2015-09-16T09:46:03.470

4

We have $| z-1| = 1$, ie

$$ \begin{align} &(z-1)(\bar{z}-1) = 1 \\ \\ \implies &z\bar{z} - z - \bar{z} = 0 \\ \\ \implies &z(\bar{z}-2) = \bar{z}(2-z) \\ \\ \implies &\overline{\left(\frac{z-2}{z}\right)} = -\left(\frac{z-2}{z}\right) \end{align} $$

i.e. $\frac{z-2}{z}$ is purely imaginary.

edited Sep 16 '15 at 09:46

answered Sep 16 '15 at 09:40

Sam Cappleman-Lynes

1,425

Verrry nice... +1. – Did Sep 16 '15 at 09:43
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Don't use \mid for the absolute value (the spacing is wrong). Just write | or if you are pedantic: \lvert .. \rvert – mrf Sep 16 '15 at 09:45
You're welcome. \mid is for set-builder notation: ${ x \mid x > 0 }$. – mrf Sep 16 '15 at 09:49
Simply brilliant +1. Like your answer better than mine – Shailesh Sep 30 '15 at 05:25

score 2 · Accepted Answer · answered Sep 16 '15 at 10:32

Look at it geometrically. Let's call A as (2,0) and O as (0,0). Let the point z be P. Then consider $\frac{z-2}{z} = \frac{2-z}{0-z}$. The argument represents the angle between PA and PO. Now because $|(z-1)| = 1$, P($z$) lies on the circle with A and O as the diameter. Hence the argument in question is a right angle. I have purposely put $\frac{z-2}{z} = \frac{2-z}{0-z}$ in a more recognizable form, given in general texts as $\frac{z_3-z_1}{z_2- z_1}$

score 0 · Answer 3 · answered Mar 24 '19 at 00:50

0

Let $z = x + iy$. Then $|z - 1|^2 = 1 \rightarrow (x-1)^2 + y^2 = 1$.

Then show that the real part of $\frac{z-2}{z}$ is $x(x - 2) + y^2$, which is zero, from above.

answered Mar 24 '19 at 00:50

1123581321

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If $z$ lies on the circle $|z-1|=1$ then$\frac {z-2} z$ is a purely imaginary number

3 Answers3