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Given $|\omega| < 1$, $\omega \neq 0$ and $|z| < 1$. Prove inequality: $$\frac{|\frac{|\omega|}{\omega}z+1|}{|1-z \bar \omega|} \le \frac{2}{1-|z|}$$ It is simple but i have problems with it. Thanks :)

Makarios
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3 Answers3

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Hint:

Triangle inequality ($|x+y|\leq |x|+|y|$) and triangle inequality ($|x-y|\geq |x|-|y|$).

P..
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Triangle inequality 1: $$ | \frac{|\omega|}{\omega}z+1|\leq |z|+1 < 2. $$ Triangle inequality 2: $$ |1-z\bar{\omega}| \geq 1 -|z||\bar{\omega}|\geq 1-|z|. $$

Julien
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Let $w=R(\cos \beta+i\sin\beta),z=r(\cos \alpha+i\sin\alpha)$

So, $$\frac{|\frac{|\omega|}{\omega}z+1|}{|1-z \bar \omega|}=\sqrt{\frac{r^2+1+2r\cos(\alpha-\beta)}{R^2r^2+1-2Rr\cos(\alpha-\beta)}}\le \sqrt{\frac{(r+1)^2}{(1-rR)^2}}$$ as $-1\le\cos(\alpha-\beta)\le 1, r^2+1+2r\cos(\alpha-\beta)\le (r+1)^2$ and $R^2r^2+1-2Rr\cos(\alpha-\beta)\ge (1-Rr)^2$

$$\frac{|\frac{|\omega|}{\omega}z+1|}{|1-z \bar \omega|}\le\frac{r+1}{1-Rr}\le \frac2{1-R}$$ as $r<1$

lab bhattacharjee
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  • Thanks, your solution is good too:) – Makarios Jan 13 '13 at 18:03
  • @Makarios, I've actually proved what the other answers assume – lab bhattacharjee Jan 13 '13 at 18:06
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    Being explicit with sines and cosines makes things overly complicated. I advise using the reverse triangle inequality $|u-v| \geq ||u| - |v||$ in the complex numbers to get upper bounds on ratios. For this problem, $||\omega|/\omega| = 1$, so numerator has abs. value at most $|z| + 1 < 2$ by the usual triangle inequality, and for denominator, from $|z\overline{\omega}| = |z||\omega| < 1$ we get $|1 - z\overline{\omega}| \geq |1 - |z\overline{\omega}|| = 1 - |z\overline{\omega}| > 1 - |z|$. Thus the left side is less than $2/(1 - |z|)$. – KCd Jan 13 '13 at 18:29
  • See my answer at http://math.stackexchange.com/questions/275547/bag-of-tricks-in-advanced-calculus-real-analysis-complex-analysis/275616#275616 for a place where the reverse triangle inequality $|u-v| \geq ||u|-|v||$ has an important role in complex analysis. – KCd Jan 13 '13 at 18:30