Assuming either |z|=1 or |w|=1... basic complex analysis

Question

Assuming either $|z|=1$ or $|w|=1$ and $\bar z w$ $\neq 1$, prove that

|$\frac{z-w}{1-\bar zw}$| $=1$

Any hint? I don't have a clue

Answer 1

Hint: If $|z|=1$, then $\overline z=1/z$. Use this to show that $|z-w|=|1-\overline zw|$.

So, $$|1-\overline zw|=\left|1-\frac wz\right|=\left|\frac 1z\right||z-w|=|z-w|.$$

Answer 2

Consider $z=x+iy$ and $w=a+bi$

Given $|z| = x^2 + y^2 = 1$ and $|w| = a^2+b^2 = 1$

Hence

$$\left|\frac{z-w}{1-\bar{z}w}\right| = 1$$

$$ \left|z-w\right| = \left|1-\bar{z}w\right| $$

Substituting in $z=x+iy$ , $w=a+bi$ and $\bar{z} = x-iy$

$$ \left|x+iy-(a+bi)\right| = \left|1-((x-iy)(a+bi))\right| $$

$$ \left|x+iy-a-bi)\right| = \left|1-(ax+by+bxi-ayi)\right| $$

$$ \left|x+iy-a-bi\right| = \left|1-ax-by-bxi+ayi)\right| $$

Grouping Imaginary and Real

$$ \left|(x-a)+i(y-b)\right| = \left|(1-ax-by)+i(ay-bx))\right| $$

$$\sqrt{(x-a)^2+(y-b)^2} = \sqrt{(1-ax-by)^2 + (ay-bx)^2}$$

$$ (x-a)^2+(y-b)^2 = (1-ax-by)^2 + (ay-bx)^2 $$

$$ x^2 - 2ax + a^2 + y^2 - 2by + b^2 = a^2x^2 + 2abxy - 2ax + b^2y^2 - 2by + 1 + a^2y^2 -2abxy +b^2x^2 $$

Simplifying

$$ (x^2 + y^2) + (a^2+b^2) - 2ax - 2by = a^2x^2+ a^2y^2 + b^2y^2+b^2x^2 - 2ax - 2by + 1 $$

$$ (x^2 + y^2) + (a^2+b^2) - 2ax - 2by = a^2(x^2+y^2) + b^2(x^2+y^2) - 2ax - 2by + 1 $$

$$ (x^2 + y^2) + (a^2+b^2) - 2ax - 2by = (a^2+b^2)(x^2+y^2) - 2ax - 2by + 1 $$

Now using $x^2+y^2=1$ and $a^2+b^2=1$

$$ (1) + (1) - 2ax - 2by = (1)(1) - 2ax + - 2by + 1 $$

$$ 2- 2ax - 2by = 2 - 2ax + - 2by $$

Hence $LHS$ = $RHS$