# Proving two Complexes' Numbers Properties

I'm having problem working with complex number on this question and was wondering if someone can walk through with me their reasoning on how to solve this/these types of questions. Thanks in advance!

Let $x, y$ be any complex numbers.

a) Prove that

$$\left|1-x \overline{y}\right|^2 - \left|x-y\right|^2 = (1-\left|x\right|^2)(1-\left|y\right|^2)$$

b) Use (a) to prove that if $\left|x\right|<1$ and $\left|y\right|<1$, then $\left|1-x \overline{y}\right|\neq0$ and we have:

$$\left|\frac{x-y}{1-x \overline{y}}\right| < 1$$

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did you start by taking $x=a+ib\;\;\;\;,y=c+id$? If you did, and if you stuck somewhere, it's wiser to include in question. – user45099 Mar 26 '13 at 23:05

Hint:

$$\left|1-x \overline{y}\right|^2 - \left|x-y\right|^2=(1-x\bar{y})\overline{(1-x\bar{y})}-(x-y)\overline{(x-y)}=(1-x\bar{y})(1-\bar{x}y)-(x-y)(\bar{x}-\bar{y})$$

Now, multiply, group and you are done....

For $b$ the Hint is that $|x|<1$ means $1-|x|^2 >0$.

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Hints:

a) Use that $|z|^2=z\cdot \bar z$ and that $\overline{z\cdot w}=\bar z\cdot\bar w$.

b) If $\ |x|,\,|y|<1$, then the right hand side of the equation of a) is positive, so $|1-x\bar y|^2>|x-y|^2$.

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