-1
$\begingroup$

Prove that if $z \ne 1$ then $\mid\dfrac{1-z}{\bar{z}-1}\mid = 1$

I have first re-written this as:

$$\frac{\mid 1 - x -iy \mid}{\mid x - iy - 1 \mid}$$

Now I am stuck. Any ideas?

  • 0
    Group together the real and imaginary parts of top and bottom, and use the formula $|a + ib|^2 = a^2 + b^2$. – 2017-02-17
  • 0
    "Then...." **what**?? – 2017-02-17
  • 0
    You should insert $=1$ immediately after the absolute value after "then" in your post. This will turn it into a statement to be shown. – 2017-02-17
  • 0
    Yes, im sorry about that!! – 2017-02-17

2 Answers 2

1

Hint: for any complex $w$ we have $|\overline w|=|w|$.

This will make it much easier than expanding in terms of $x+iy$.

  • 0
    Ok but is $\mid{1-z}\mid = \mid \bar{z} - 1 \mid$ ?? – 2017-02-17
  • 0
    Choose a suitable expression for $w$ in my hint. – 2017-02-17
0

Well, if $\space\exists\space\text{z}\in\mathbb{C}$ and $\text{z}\ne1$:

$$\mathcal{Z}=\left|\frac{1-\text{z}}{\overline{\text{z}}-1}\right|=\frac{\left|1-\text{z}\right|}{\left|\overline{\text{z}}-1\right|}=\frac{\left|1-\Re\left(\text{z}\right)-\Im\left(\text{z}\right)i\right|}{\left|\Re\left(\text{z}\right)-\Im\left(\text{z}\right)i-1\right|}=\frac{\sqrt{\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)}}{\sqrt{\left(\Re\left(\text{z}\right)-1\right)^2+\Im^2\left(\text{z}\right)}}\tag1$$

Now, we can write:

$$\mathcal{Z}=\sqrt{\frac{\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)}{\left(\Re\left(\text{z}\right)-1\right)^2+\Im^2\left(\text{z}\right)}}=\sqrt{\frac{\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)}{\left(1-\Re\left(\text{z}\right)\right)^2+\Im^2\left(\text{z}\right)}}=\sqrt{1}=1\tag2$$

Because:

$$\left(1-\Re\left(\text{z}\right)\right)^2=\left(\Re\left(\text{z}\right)-1\right)^2\tag3$$