Show that $$|z| \lt 1 \Rightarrow |z-i| \lt \sqrt 2$$ $x^2+y^2 \lt 1$ How can I show $x^2+(y-1)^2 \lt 2$ ? I’m sorry, i know it’s so easy but I couldnt obtain it in no way.

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    $\begingroup$ Not true at all, this "so easy inequality", that's why you are struggling. $\endgroup$ – Teresa Lisbon Apr 5 '19 at 9:06
  • $\begingroup$ You have probably a typo. Instead of $\sqrt 2$ put $2$. $\endgroup$ – user555729 Apr 5 '19 at 9:09
  • $\begingroup$ @астонвіллаолофмэллбэрг could you please glance my comment below? $\endgroup$ – user519955 Apr 5 '19 at 9:10

This is incorrect. Consider $z=-\frac{1}{2}i$. Then $|z|=\frac{1}{2}<1$ but $|z-i|=\left|-\frac{1}{2}i-i\right|=\left|-\frac{3}{2}i\right|=\frac{3}{2}>\sqrt{2}$.

  • $\begingroup$ OK. Thanks. How can I show |z-i|<sqrt2 when finding taylor expansion around i of 1/z via using Maclaurin expansion of 1/z? $\endgroup$ – user519955 Apr 5 '19 at 9:09
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    $\begingroup$ @user519955 You cannot show $|z-i| < \sqrt 2$ because it is not true, right? Please clarify your context further, maybe we can infer the right question from this. $\endgroup$ – Teresa Lisbon Apr 5 '19 at 9:31

After the correction $\sqrt 2\to 2$: use the triangle inequality: $$|z - i|\le |z| + |i|< 1 + 1 = 2.$$


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