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:

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

  • 2
    $\begingroup$ 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. $\endgroup$ – user45099 Mar 26 '13 at 23:05


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

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

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



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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