3

Given $z \ne -1$. Prove that $(z-1) \over (z+1)$ is an Imaginary number if and only if $|z| = 1$.

I tried computing $(z-1) \over (z+1)$ by multiplying like that: $(z-1) \over (z+1)$$(z-1) \over (z-1)$ and then getting: $(z-1)^2$

but if $z = 1$ then it's zero, and it's not an imaginary number. Any suggestions?

Martin Sleziak
  • 53,687
Ilan Aizelman WS
  • 4,128
  • Maybe its like $2Im(z-1)^2$ something.. which means it always an Imaginary? I'm not quite sure :( – Ilan Aizelman WS Jun 29 '14 at 15:44
  • 1
    You can regard $0=0i$ as an imaginary number in this case. It lies in the intersection of the real axis and the imaginary axis, so it is both real and pure imaginary. It really comes down to how you define pure imaginary -- if you just require that the real part vanishes, this is okay. – MPW Jun 29 '14 at 15:58
  • I just got it with $(z-1)(\overline{(z+1)} = (z-1)(\overline{(z)}+\overline{(1))} = (z-1)\overline{(z}+1)$ – Ilan Aizelman WS Jun 29 '14 at 16:06

2 Answers2

5

I believe you have meant purely imaginary

Setting $z=a+ib$ where $a,b$ are real

$$\frac{z-1}{z+1}=\frac{a+ib-1}{a+ib+1}=\frac{(a-1+ib)(a+1-ib)}{(a+1)^2+b^2}$$

$$=\frac{(a)^2-(1-ib)^2}{(a+1)^2+b^2}=\frac{a^2+b^2-1+2ib}{(a+1)^2+b^2}$$

lab bhattacharjee
  • 274,582
5

Multiply numerator and denominator by $z^*+1$:

$$\frac{z-1}{z+1}=\frac{(z-1)(z^*+1)}{|z+1|^2}=\frac{zz^*+z-z^*-1}{|z+1|^2}= \frac{|z|^2-1+2iIm\{z\}}{|z+1|^2}$$

Obviously, the denominator is real, and so this number is purely imaginary if $|z|^2-1=0$, i.e. if $|z|=1$.

Matt L.
  • 10,636